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HDU2934�Q�Gargoyle

LLawliet — Sat, 15 Oct 2011 14:16:00 GMT

Gargoyle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 139 Accepted Submission(s): 22

Problem Description

Gargoyles can trace their history back many thousands of years to ancient Egypt, Greece, and Rome. Terra cotta waterspouts were formed in the shapes of animals such as lions and birds to serve the physical function of running the
rainwater away from the walls and foundations of buildings, and the spiritual function of protecting from evil forces.
Have you ever dreamed of creating your own castle with a lot of beautiful gargoyles on the walls? To your knowledge,
the speed of water coming out of each gargoyle should be identical, so an elaborately designed water system is required.
The water system consists of a huge reservoir and several interconnecting water pipes. Pipes cannot save water, so the total incoming and outgoing speed of water should be equal at each connection.
All the water from gargoyles flows into the reservoir, which is located at the bottom of the castle. Some pipes are connecting the reservoir, but water can only go from the reservoir to pipes, but never from pipes back to the reservoir. A micro-processor is installed inside each pipe, so the speed of water could easily be controlled. However, the microprocessors consume electricity. The exact cost in each pipe is proportional to the speed of water. If the cost constant in the i-th pipe is ci, the electricity cost in that pipe is civi, where vi is the speed of water in that pipe. Write a program to find the optimal configuration of the water system (i.e. the water speed in each pipe) of your dream castle, so that the total cost is minimized. It is always possible to build a water system.

Input

The input consists of several test cases. The first line of each case contains three integers n, m and k (1 ≤ n ≤ 25, 1 ≤ m ≤ 50, 1 ≤ k ≤ 1000), the number of gargoyles, the number of pipe connections and the number of pipes. The following k lines each contains five integers a, b, l, u, c (0 ≤ a, b ≤ n + m, 0 ≤ l ≤ u ≤ 100, 1 ≤ c ≤ 100), describing each pipe. a and b
are the incoming and outgoing vertex number (reservoir is 0, gargoyles are numbered 1 to n, pipe connections are numbered n + 1 to n + m), lower-bound and upper-bound of water speed, and the cost constant. No pipe connects two identical vertices. For every pipe, the incoming vertex will never be a gargoyle, and the outgoing vertex will never be the reservoir. For every pair of vertices, there could be at most one pipe connecting them (if a pipe is going from a to b, no pipes can go from a to b, or from b to a). The last test case is followed by a single zero, which should not be processed.

Output

For each test case, print the case number and minimal cost to two decimal places.

Sample Input

Sample Output

Case 1: 60.00

Source

2006 Asia Regional Xi-An

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lcy

06�q�亚�z�区域赛西安赛区的一道网�l�流�Q�相当犀�?.....我WA很久很久才勉��?......

一�Q��M��分析

本题�l�出了一个网�l�流的模型。以�U�水池�ؓ源点�Q�每个连接点和喷水口为结点，另外建立一个超�U�汇�Q�每个喷水口向超�U�汇�q�一条边。题目就是求一个有上下界的最��费用流�Q�其中到汇点的每一条边的流量必��ȝ��{?*)�?/div>

很多文献�Ҏ��上下界的费用��进行了讨论�Q�本文不再赘�q�。本文主要讨��军_��何解�?*)的要求�?br />

二．判断可行��?/div>

我们不妨设所有连接到汇点的边的流量均为F�Q�我们要做的�W�一步就是：判断是否存在�q�样一�l�可行流使得F = F0成立�?/div>

我们不妨�l�每一条连接到汇点的边建立上下界[0�Q�F0]。我们给��Z��下引理：

引理一�Q�存在满��得F0的可行流�Q�当且仅当该�|�络的最大流为F0N�?/div>

引理二：若最大流��于F0N�Q�则不存在F >= F0的可行流�Q?/div>

引理三：若不存在可行��，则不存在F <= F0的可行流�Q?/div>

由引理二、三可知�Q?/div>

引理四：

存在可行��的F的定义域��Z��D�连�l�的区间[low, top]�?/div>

�Ҏ��引理二、三和四�Q�我们不难��用二分法扑և��q�一�D�F的可行区间[low, top]�?/div>

三．扑և�最��费用流

记cost(F)表示在F成立的最��费用。我们猜��，cost(F)与F成正比！

感觉上，F��大�Q�流量越大，那么每条边上消耗的费用也就��大�Q�所以猜惛_��该成立。可惜的是，按照该想法提交是无法通过��试数据的。因为：

大家不免疑问�Q�单调函��C��也是凸的么？

注意到有上下界的最��费用由两部分构成：附加�|�络的最��费用及�D�余�|�络的最��费用。注意到前者是关于F不增的后者是关于F不减的，如果前者的递减速度大于后者的递增速度�Q�那么cost(F)��成��Z��凸函数�?/div>

四．��法��程

�׃��最��费用流的时间复杂度不好分析�Q�本文直接认为是O(kN2)�Q�其中k表示增广�ơ数�Q�算法�ȝ��旉��复杂度�ؓO(kN2logC)�?/div>

LLawliet 2011-10-15 22:16 发表评论

POJ1149:PIGS

LLawliet — Sat, 15 Oct 2011 14:12:00 GMT

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10566		Accepted: 4622

Description

Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs.
All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold.
More precisely, the procedure is as following: the customer arives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.
An unlimited number of pigs can be placed in every pig-house.
Write a program that will find the maximum number of pigs that he can sell on that day.

Input

The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N.
The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000.
The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line):
A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.

Output

The first and only line of the output should contain the number of sold pigs.

Sample Input

Sample Output

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#include<stdio.h>

#include<string.h>
#include<iostream>
#include<queue>
#define N 1010
using namespace std;
const int maxnode=60000;
const int maxedge=320000;
const int  inf = 1 << 30;
int S,T,cnt,n,m;
int head[maxnode],gap[maxnode],pre[maxnode],cur[maxnode],dis[maxnode];
struct Edge
{
    int S,t;
    int next;
    int w;
}st[maxedge];
void init()
{
    memset(head,-1,sizeof(head));
    cnt=0;
}
void AddEdge(int S,int t,int w)
{
    st[cnt].S=S;
    st[cnt].t=t;
    st[cnt].w=w;
    st[cnt].next=head[S];
    head[S]=cnt;
    cnt++;
    st[cnt].S=t;
    st[cnt].t=S;
    st[cnt].w=0;
    st[cnt].next=head[t];
    head[t]=cnt;
    cnt++;
}
void bfs()
{
    memset(gap,0,sizeof(gap));
    memset(dis,-1,sizeof(dis));
    queue<int >Q;
    Q.push(T);
    dis[T]=0;
    gap[0]=1;
    int k,t;
    while(!Q.empty())
    {
        k=Q.front();
        Q.pop();
        for(int i=head[k];i!=-1;i=st[i].next)
        {
            t=st[i].t;
            if(dis[t]==-1&&st[i^1].w>0)
            {
                dis[t]=dis[k]+1;
                gap[dis[t]]++;
                Q.push(t);
            }
        }
    }
}
int sap()
{
    int i;
    for( i=S;i<=T;i++)cur[i]=head[i];
    pre[S]=S;
    int u=S,v;
    int  flow=0;
    int aug=inf;
    bool flag;
    while(dis[S]<=T)
    {
        flag=false;
        for( i=cur[u];i!=-1;i=st[i].next)
        {
            v=st[i].t;
            if(st[i].w>0&&dis[u]==dis[v]+1)
            {
                cur[u]=i;
                flag=true;
                pre[v]=u;
                aug=(aug>st[i].w)?st[i].w:aug;
                u=v;
                if(v==T)
                {
                    flow+=aug;
                    for(u=pre[u];v!=S;u=pre[u])
                    {
                        v=u;
                        st[cur[u]].w-=aug;
                        st[cur[u]^1].w+=aug;
                    }
                    aug=inf;
                }
                break;
            }
        }
        if(flag==true)continue;
        int mint=T;
        for( i=head[u];i!=-1;i=st[i].next)
        {
            v=st[i].t;
            if(st[i].w>0&&mint>dis[v])
            {
                cur[u]=i;
                mint=dis[v];
            }
        }
        gap[dis[u]]--;
        if(gap[dis[u]]==0)break;
        gap[dis[u]=mint+1]++;
        u=pre[u];
        if(u==S)aug=inf;
    }
    return flow;
}

int main()
{
    int map[N][N];
    int num[N];
    int f[N];
    while (scanf("%d%d", &m, &n) != EOF)
    {
        memset(map, 0, sizeof(map));
        memset(f, 0, sizeof(f));
        init();
        S = 0;
        T = n + 1;
        for (int i = 1; i <= m; ++i)
            scanf("%d", &num[i]);
        for (int i = 1; i <= n; ++i)
        {
            int a, b;
            scanf("%d", &a);
            for (int j = 1; j <= a; ++j)
            {
                int x;
                scanf("%d", &x);
                if (f[x] == 0)
                {
                    map[S][i] += num[x];
                    f[x] = i;
                }
                else
                {
                    map[f[x]][i] = inf;
                    f[x] = i;
                }
            }
            scanf("%d", &b);
            map[i][T] += b;
        }
        for (int i = S; i <= T; ++i)
            for (int j = S; j <= T; ++j)
            if (map[i][j])
            {
                //printf("%d %d %d\n", i, j, map[i][j]);
                AddEdge(i, j, map[i][j]);
            }
        bfs();
        printf("%d\n", sap());
    }
}

/*
4 2
1 2 3 5
3 1 2 3 10
1 4 5

*/

LLawliet 2011-10-15 22:12 发表评论

HDU3879:Base Station

LLawliet — Sat, 15 Oct 2011 14:10:00 GMT

Base Station

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 65768/32768 K (Java/Others)
Total Submission(s): 844 Accepted Submission(s): 353

Problem Description

A famous mobile communication company is planning to build a new set of base stations. According to the previous investigation, n places are chosen as the possible new locations to build those new stations. However, the condition of each position varies much, so the costs to built a station at different places are different. The cost to build a new station at the ith place is P_i (1<=i<=n).

When complete building, two places which both have stations can communicate with each other.

Besides, according to the marketing department, the company has received m requirements. The ith requirement is represented by three integers A_i, B_i and C_i, which means if place A_iand B_i can communicate with each other, the company will get C_i profit.

Now, the company wants to maximize the profits, so maybe just part of the possible locations will be chosen to build new stations. The boss wants to know the maximum profits.

Input

Multiple test cases (no more than 20), for each test case:
The first line has two integers n (0The second line has n integers, P1 through Pn, describes the cost of each location.
Next m line, each line contains three integers, A_i, B_i and C_i, describes the ith requirement.

Output

One integer each case, the maximum profit of the company.

Sample Input

Sample Output

Author

liulibo

Source

2011 Multi-University Training Contest 5 - Host by BNU

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lcy

论文题，Amber最��割模型里面的，最大权闭合图，因�ؓ数组开的太大了�Q�吃了一�ơRE......
最大权闭合图不用说了，边看成收益点�Q�连向S�Q�流量是�Ҏ��Q�站点看成花费点�Q�连向T�Q�流量也是点权，其他按照原图�q�边�Q�流量是无限大，之后做一�ơ最��割�Q�割值就是你未选的收益�?选定��p��点（因�ؓ是闭合图�Q�所以割肯定是简单割�Q�想一下割的定义，��明白割值的含义了）�Q�用你��L��?割��|��是�{�案�?br />用SAP求的最��割�Q�渐渐爱上SAP了，Dinic不用�?....
代码�Q?br />

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;

const int maxnode = 60000;
const int maxedge = 320000;
const long long inf = (1LL << 35);

int S, T, cnt;
int head[maxnode], gap[maxnode], pre[maxnode], cur[maxnode], dis[maxnode];

struct Edge
{
    int s, t;
    int next;
    long long w;
} st[maxedge];

void init()
{
    memset(head, -1, sizeof(head));
    cnt = 0;
}

void AddEdge(int s, int t, long long w)
{
    st[cnt].s = s;
    st[cnt].t = t;
    st[cnt].w = w;
    st[cnt].next = head[s];
    head[s] = cnt;
    cnt++;

    st[cnt].s = t;
    st[cnt].t = s;
    st[cnt].w = 0;
    st[cnt].next = head[t];
    head[t] = cnt;
    cnt++;
}

void bfs()
{
    memset(gap, 0, sizeof(gap));
    memset(dis, -1, sizeof(dis));
    queue <int> Q;
    Q.push(T);
    dis[T] = 0;
    gap[0] = 1;
    int k, t;
    while (!Q.empty())
    {
        k = Q.front();
        Q.pop();
        for (int i = head[k]; i != -1; i =st[i].next)
        {
            t = st[i].t;
            if (dis[t] == -1 && st[i ^ 1].w > 0)
            {
                dis[t] = dis[k] + 1;
                gap[dis[t]]++;
                Q.push(t);
            }
        }
    }
}

long long sap()
{
    int i;
    for (i = S; i <= T; ++i)
        cur[i] = head[i];
    pre[S] = S;
    int u = S, v;
    long long flow = 0;
    long long aug = inf;
    bool flag;
    while (dis[S] <= T)
    {
        flag = false;
        for (i = cur[u]; i != -1; i = st[i].next)
        {
            v = st[i].t;
            if (st[i].w > 0 && dis[u] == dis[v] + 1)
            {
                cur[u] = i;
                flag = true;
                pre[v] = u;
                aug = (aug > st[i].w) ? st[i].w : aug;
                u = v;
                if (v == T)
                {
                    flow += aug;
                    for (u = pre[u]; v != S; u = pre[u])
                    {
                        v = u;
                        st[cur[u]].w -= aug;
                        st[cur[u] ^ 1].w += aug;
                    }
                    aug = inf;
                }
                break;
            }
        }
        if (flag == true) continue;
        int mint = T;
        for (i = head[u]; i != -1; i = st[i].next)
        {
            v = st[i].t;
            if (st[i].w > 0 && mint > dis[v])
            {
                cur[u] = i;
                mint = dis[v];
            }
        }
        gap[dis[u]]--;
        if (gap[dis[u]] == 0) break;
        gap[dis[u] = mint + 1]++;
        u = pre[u];
        if (u == S) aug = inf;
    }
    return flow;
}

int main()
{
    int n, m;
    while (scanf("%d%d", &n, &m) != EOF)
    {
        init();
        S = 0;
        T = n + m + 1;
        int sum = 0;
        for (int i = 1; i <= n; ++i)
        {
            int x;
            scanf("%d", &x);
            AddEdge(m + i, T, x);
        }
        for (int i = 1; i <= m; ++i)
        {
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            AddEdge(S, i, c);
            AddEdge(i, m + a, inf);
            AddEdge(i, m + b, inf);
            sum += c;
        }
        bfs();
        sum -= sap();
        printf("%d\n", sum);
    }
    return 0;
}

LLawliet 2011-10-15 22:10 发表评论

POJ2914:Minimum Cut

LLawliet — Sat, 15 Oct 2011 14:10:00 GMT

Minimum Cut

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 5403		Accepted: 2113
Case Time Limit: 5000MS

Description

Given an undirected graph, in which two vertices can be connected by multiple edges, what is the size of the minimum cut of the graph? i.e. how many edges must be removed at least to disconnect the graph into two subgraphs?

Input

Input contains multiple test cases. Each test case starts with two integers N and M (2 ≤ N ≤ 500, 0 ≤ M ≤ N × (N − 1) ⁄ 2) in one line, where N is the number of vertices. Following are M lines, each line contains M integers A, B and C (0 ≤ A, B < N, A ≠ B, C > 0), meaning that there C edges connecting vertices A and B.

Output

There is only one line for each test case, which contains the size of the minimum cut of the graph. If the graph is disconnected, print 0.

Sample Input

Sample Output

2
1
2

Source

Baidu Star 2006 Semifinal
Wang, Ying (Originator)
Chen, Shixi (Test cases)

全局最��割�Q�模杉K��Q�Stor-Wagner��法�Q�大概就是prime的最大生成树�Q�具体的�|�上有很多，可以google到的�Q�不多说了�?br />代码�Q?/div>

#include<stdio.h>

#include<string.h>

#define NN 504
#define INF 1 << 30
int vis[NN];
int wet[NN];
int combine[NN];
int map[NN][NN];

int S, T, minCut, N;
void Search()
{
    int i, j, Max, tmp;
    memset(vis, 0, sizeof(vis));
    memset(wet, 0, sizeof(wet));
    S = T = -1;
    for (i = 0; i < N; i++)
    {
        Max = -INF;
        for (j = 0; j < N; j++)
        {
            if (!combine[j] && !vis[j] && wet[j] > Max)
            {
                tmp = j;
                Max = wet[j];
            }
        }
        if (T == tmp) return;
        S = T;
        T = tmp;
        minCut = Max;
        vis[tmp] = 1;
        for (j = 0; j < N; j++)
        {
            if (!combine[j] && !vis[j])
            {
                wet[j] += map[tmp][j];
            }
        }
    }
}
int Stoer_Wagner()
{
    int i, j;
    memset(combine, 0, sizeof(combine));
    int ans = INF;
    for (i = 0; i < N - 1; i++)
    {
        Search();
        if (minCut < ans) ans = minCut;
        if (ans == 0) return 0;
        combine[T] = 1;
        for (j = 0; j < N; j++)
        {
            if (!combine[j])
            {
                map[S][j] += map[T][j];
                map[j][S] += map[j][T];
            }
        }
    }
    return ans;
}
int main()
{
    int a, b, c, M;
    while(scanf("%d%d", &N, &M) != EOF)
    {
        memset(map, 0, sizeof(map));
        while(M--)
        {
            scanf("%d%d%d", &a, &b, &c);
            map[a][b] += c;
            map[b][a] += c;
        }
        printf("%d\n", Stoer_Wagner());
    }
    return 0;
}

LLawliet 2011-10-15 22:10 发表评论

POJ2125:Destroying The Graph

LLawliet — Sat, 15 Oct 2011 14:09:00 GMT

Destroying The Graph

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4718		Accepted: 1436		Special Judge

Description

Alice and Bob play the following game. First, Alice draws some directed graph with N vertices and M arcs. After that Bob tries to destroy it. In a move he may take any vertex of the graph and remove either all arcs incoming into this vertex, or all arcs outgoing from this vertex.
Alice assigns two costs to each vertex: Wi⁺ and Wi^-. If Bob removes all arcs incoming into the i-th vertex he pays Wi⁺ dollars to Alice, and if he removes outgoing arcs he pays Wi^- dollars.
Find out what minimal sum Bob needs to remove all arcs from the graph.

Input

Input file describes the graph Alice has drawn. The first line of the input file contains N and M (1 <= N <= 100, 1 <= M <= 5000). The second line contains N integer numbers specifying Wi⁺. The third line defines Wi^- in a similar way. All costs are positive and do not exceed 10⁶ . Each of the following M lines contains two integers describing the corresponding arc of the graph. Graph may contain loops and parallel arcs.

Output

On the first line of the output file print W --- the minimal sum Bob must have to remove all arcs from the graph. On the second line print K --- the number of moves Bob needs to do it. After that print K lines that describe Bob's moves. Each line must first contain the number of the vertex and then '+' or '-' character, separated by one space. Character '+' means that Bob removes all arcs incoming into the specified vertex and '-' that Bob removes all arcs outgoing from the specified vertex.

Sample Input

Sample Output

Source

Northeastern Europe 2003, Northern Subregion

一道典型的最��点权覆盖问题，SAP速度很好看，之后需要搜索一下用�q�的点，输出卛_��?br />代码�Q?br />

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define N 10010
#define M 20010
#define inf 1 << 30
#define eps 1 << 29
using namespace std;

int mark[N];
int cnt, n, m, s, t;
int head[N];
int NN;

struct edge
{
    int v, next, w;
} edge[M];

void addedge(int u, int v, int w)
{
    edge[cnt].v = v;
    edge[cnt].w = w;
    edge[cnt].next = head[u];
    head[u] = cnt++;
    edge[cnt].v = u;
    edge[cnt].w = 0;
    edge[cnt].next = head[v];
    head[v] = cnt++;
}

int sap()
{
    int pre[N], cur[N], dis[N], gap[N];
    int flow = 0, aug = inf, u;
    bool flag;
    for (int i = 1; i <= NN; ++i)
    {
        cur[i] = head[i];
        gap[i] = dis[i] = 0;
    }
    gap[s] = NN;
    u = pre[s] = s;
    while (dis[s] < NN)
    {
        flag = 0;
        for (int &j = cur[u]; j != -1; j = edge[j].next)
        {
            int v = edge[j].v;
            if (edge[j].w > 0 && dis[u] == dis[v] + 1)
            {
                flag = 1;
                if (edge[j].w < aug) aug = edge[j].w;
                pre[v] = u;
                u = v;
                if (u == t)
                {
                    flow += aug;
                    while (u != s)
                    {
                        u = pre[u];
                        edge[cur[u]].w -= aug;
                        edge[cur[u] ^ 1].w += aug;
                    }
                    aug = inf;
                }
                break;
            }
        }
        if (flag)
            continue;
        int mindis = NN;
        for (int j = head[u]; j != -1; j = edge[j].next)
        {
            int v = edge[j].v;
            if (edge[j].w > 0 && dis[v] < mindis)
            {
                mindis = dis[v];
                cur[u] = j;
            }
        }
        if ((--gap[dis[u]]) == 0)
            break;
        gap[dis[u] = mindis + 1]++;
        u = pre[u];
    }
    return flow;
}

void init()
{
    cnt = 0;
    memset(head, -1, sizeof(head));
    memset(mark, 0, sizeof(mark));
}

void dfs(int x) //不同于单�U�的SAP�Q�加入了一个搜素点集元素的函数�Q�通过head数组的记录信息搜索�?br />{
    mark[x] = 1;
    for (int i = head[x]; i; i = edge[i].next)
    {
        if (edge[i].w > 0 && !mark[edge[i].v]) dfs(edge[i].v);
    }
}

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        init();
        int wp[101], wm[101], w, len = 1, ans[105];
        s = 0;
        t = 2 * n + 1;
        NN = 2 * n + 2;
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &wp[i]);
            addedge(i + n, t, wp[i]);
        }
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &wm[i]);
            addedge(s, i, wm[i]);
        }
        for (int i = 1; i <= m; ++i)
        {
            int x, y;
            scanf("%d%d", &x, &y);
            addedge(x, y + n, inf);
        };
        w = sap();
        dfs(s);
        for (int i = 1; i <= n; ++i)
        {
            if (!mark[i]) ans[len++] = i;
            if (mark[i + n]) ans[len++] = i + n;
        }
        len--;
        printf("%d\n%d\n", w, len);
        for (int i = 1; i <= len; ++i)
        {
            if (ans[i] <=n) printf("%d -\n", ans[i]);
            else printf("%d +\n", ans[i] - n);
        }
    }
    return 0;
}

LLawliet 2011-10-15 22:09 发表评论