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reiks — Thu, 19 May 2011 05:18:00 GMT

�E�序设计领域里，每个人都想飞�?br />但是�Q�还没学会走之前�Q�连跑都别想�Q?br />
勿在��沙�{�高��|��

从今天开始，�t�实的学习，不在��躁�Q�加油！

reiks 2011-05-19 13:18 发表评论

最大流 Edmonds-Karp

reiks — Sat, 29 Aug 2009 05:39:00 GMT

//Edmonds-Karp
//return the largest flow;flow[] will record every edge's flow
//n, the number of nodes in the graph;cap, the capacity
//O(VE^2)
#define N 100
#define inf 0x3f3f3f3f
int Edmonds_Karp(int n,int cap[][N],int source,int sink)
{
    int flow[N][N];
    int pre[N],que[N],d[N]; // d 是增�q��\长度�Q�pre 记录前驱�Q�que是BFS队列
    int p,q,t,i,j;
    if (source==sink) return inf;
    memset(flow,0,sizeof(flow));
    while (true)
    {
        memset(pre,-1,sizeof(pre));
        d[source]=inf;
        p=q=0, que[q++] = source;
        while(p < q&&pre[sink]<0)    // BFS 找�\�?/span>
        {
            t=que[p++];
            for (i=0;i<n;i++)
                if ( pre[i]<0 && (j=cap[t][i]-flow[t][i]) ) // j取得�D�余路径�?/span>
                    pre[que[q++] = i] = t,d[i] = min(d[t], j);
        }
        if (pre[sink]<0) break;    // 找不到增�q��\�Q�退�?/span>
        for (i=sink; i!=source; i=pre[i])
        {
            flow[pre[i]][i]+=d[sink];    // 正向��量�?/span>
            flow[i][pre[i]]-=d[sink];    // 反向��量�?/span>
        }
    }
    for (j=i=0; i<n; j+=flow[source][i++]);
    return j;
}

reiks 2009-08-29 13:39 发表评论

树状数组

reiks — Fri, 28 Aug 2009 02:35:00 GMT

#include <stdio.h>
#include <memory.h>
#define N 1000

class treearray
{
       public:
       int c[N],n;
       void clear()
       {
             memset(this,0,sizeof(*this));
       }
       int lowbit(int x)
       {
             return x&(x^(x-1));
       }
       void change(int i,int d)
       {
             for (;i<=n;i+=lowbit(i))       c[i]+=d;
       }
       int getsum(int i)
       {
             int t;
             for (t=0;i>0;i-=lowbit(i))       t+=c[i];
             return t;
       }
}t;

main()//附一个测试程�?/span>
{
       int i,x;
       t.clear();
       scanf("%d",&t.n);
       for (i=1;i<=t.n;i++)
       {
             scanf("%d",&x);
             t.change(i,x);
       }
       for (;scanf("%d",&x),x;)       printf("%d\n",t.getsum(x));
       return 0;
}

reiks 2009-08-28 10:35 发表评论

reiks — Fri, 28 Aug 2009 02:35:00 GMT

摘要: #includebool map[102][302],use[302];int link[302],n,m;bool dfs(int);int main(){ int t,v,i,j,x,num; scanf("%d",&am... 阅读全文

reiks 2009-08-28 10:35 发表评论

reiks — Fri, 28 Aug 2009 02:34:00 GMT

#define MAXSIZE 50001
int father[MAXSIZE];
int rank[MAXSIZE];

void initial()
{
    memset(rank, 0, sizeof(rank));
    for ( int i = 0; i < MAXSIZE; ++i )
        father[i] = -1;
}

int find_set(int x)
{
    int r = x, q;

    while(father[r] != -1)
    {
        r = father[r];
    }

    while(x != r)
    {
        q = father[x];
        father[x] = r;
        x = q;
    }
    return r;
}

void union_set(int x, int y)
{
    int a = find_set(x);
    int b = find_set(y);
    if (a == b)
        return;
    if (rank[a] > rank[b])
    {
        father[b] = a;
    }
    else
    {
        father[a] = b;
        if (rank[a] == rank[b])
        {
            ++rank[b];
        }
    }
}

reiks 2009-08-28 10:34 发表评论

Topsort

reiks — Fri, 28 Aug 2009 02:33:00 GMT

/**
* TOPSORT(��单版) 拓扑排序(Topological Sort)
* 输入�Q�有向图g
* 输出�Q�是否存在拓扑排序，如果存在�Q�获取拓扑排序序列seq
* �l�构�Q�图g用邻接矩阵表�C?br> * ��法�Q�广度优先搜�?BFS)
* 复杂度：O(|V|^2)
*/

#include <iostream>
#include <vector>
#include <queue>
#include <iterator>
#include <algorithm>
#include <numeric>
#include <climits>
using namespace std;

int n;                            // n �Q�顶点个�?nbsp;
vector<vector<int> > g;           // g �Q�图(graph)(用邻接矩�?adjacent matrix)表示)
vector<int> seq;                // seq �Q�拓扑序�?sequence)

bool TopSort()
{
    vector<int> inc(n, 0);
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
             if (g[i][j] < INT_MAX) ++inc[j]; // 计算每个��点的入度，
    queue<int> que;
    for (int j = 0; j < n; ++j)
        if (inc[j] == 0) que.push(j); // 如果��点的入度�ؓ0�Q�入队�?/span>
    int seqc = 0;
    seq.resize(n);
    while (!que.empty())     // 如果队列que非空�Q?/span>
    {
        int v = que.front(); que.pop();
        seq[seqc++] = v;      // ��点v出队�Q�放入seq中，
        for (int w = 0; w < n; ++w)     // 遍历所有v指向的顶点w�Q?/span>
            if (g[v][w] < INT_MAX)
                if (--inc[w] == 0) que.push(w); // 调整w的入度，如果w的入度�ؓ0�Q�入队�?nbsp;
    }
    return seqc == n; // 如果seq已处理顶�Ҏ��为n�Q�存在拓扑排序，否则存在回�\�?/span>
}

int main()
{
    n = 7;
    g.assign(n, vector<int>(n, INT_MAX));
    g[0][1] = 1, g[0][2] = 1, g[0][3] = 1;
    g[1][3] = 1, g[1][4] = 1;
    g[2][5] = 1;
    g[3][2] = 1, g[3][5] = 1, g[3][6] = 1;
    g[4][3] = 1, g[4][6] = 1;
    g[6][5] = 1;

    if (TopSort())
    {
         copy(seq.begin(), seq.end(), ostream_iterator<int>(cout, " "));
         cout << endl;
    }
    else
    {
         cout << "circles exist" << endl;
    }

    system("pause");
    return 0;
}

reiks 2009-08-28 10:33 发表评论

reiks — Fri, 28 Aug 2009 02:32:00 GMT

/*
Name: Trie树的基本实现
Author: MaiK
Description: Trie树的基本实现 ,包括查找插入和删除操�?卫星数据可以因情况而异)
*/
#include<algorithm>
#include<iostream>
using namespace std;

const int sonnum=26,base='a';
struct Trie
{
    int num;  //to remember how many word can reach here,that is to say,prefix
    bool terminal;  //If terminal==true ,the current point has no following point
    struct Trie *son[sonnum];  //the following point
};
Trie *NewTrie()// create a new node
{
    Trie *temp=new Trie;
    temp->num=1;
    temp->terminal=false;
    for (int i=0; i<sonnum; ++i)
        temp->son[i] = NULL;
    return temp;
}
void Insert(Trie *pnt,char *s,int len)// insert a new word to Trie tree
{
    Trie *temp=pnt;
    for (int i=0;i<len;++i)
    {
        if (temp->son[s[i]-base]==NULL)
            temp->son[s[i]-base]=NewTrie();
        else
            temp->son[s[i]-base]->num++;
        temp=temp->son[s[i]-base];
    }
    temp->terminal=true;
}
void Delete(Trie *pnt)  // delete the whole tree
{
    if (pnt!=NULL)
    {
        for (int i=0;i<sonnum;++i)
            if (pnt->son[i]!=NULL)
                Delete(pnt->son[i]);
        delete pnt;
        pnt=NULL;
    }
}
Trie* Find(Trie *pnt,char *s,int len)  //trie to find the current word
{
    Trie *temp=pnt;
    for (int i=0;i<len;++i)
        if (temp->son[s[i]-base]!=NULL)
            temp=temp->son[s[i]-base];
        else return NULL;
    return temp;
}

reiks 2009-08-28 10:32 发表评论

大整��C��除小整数

reiks — Fri, 28 Aug 2009 01:22:00 GMT

// 大整��C��以一个小整数
void big_mul(int d[], int s[], int n)
{
    int plus = 0;
    for (int i = 1; i < 61; ++i)
    {
        d[i] = s[i] * n;
        d[i] += plus;
        plus = d[i] / 10;
        d[i] %= 10;
    }
}

// 大整数除以一个小整数
void big_div(int d[], int s[], int n)
{
    int left = 0;
    for (int i = 60; i > 0; --i)
    {
        left *= 10;
        left += s[i];
        if (left < n)
        {
            d[i] = 0;
        }
        else
        {
            d[i] = left / n;
            left %= n;
        }
    }
}

reiks 2009-08-28 09:22 发表评论

RMQ问题ST��法

reiks — Fri, 28 Aug 2009 01:20:00 GMT

/*
RMQ(Range Minimum/Maximum Query)问题�Q?br>   RMQ问题是求�l�定区间中的最值问题。当�Ӟ��最��单的��法是O(n)的，但是对于查询�ơ数很多�Q�设�|�多�?00万次�Q�，O(n)的算法效率不够。可以用�U�段树将��法优化到O�Q�logn)�Q�在�U�段树中保存�U�段的最��|��。不�q�，Sparse_Table��法才是最好的�Q�它可以在O(nlogn)的预处理以后实现O(1)的查询效率。下面把Sparse Table��法分成预处理和查询两部分来说明(以求最��gؓ�?�?br>
预处�?
预处理��用DP的思想�Q�f(i, j)表示[i, i+2^j - 1]区间中的最��|��我们可以开辟一个数�l�专门来保存f(i, j)的倹{�?br>例如�Q�f(0, 0)表示[0,0]之间的最��?��是num[0], f(0, 2)表示[0, 3]之间的最��? f(2, 4)表示[2, 17]之间的最��?br>注意, 因�ؓf(i, j)可以由f(i, j - 1)和f(i+2^(j-1), j-1)导出, 而递推的初�?所有的f(i, 0) = i)都是已知�?br>所以我们可以采用自底向上的��法递推地给出所有符合条件的f(i, j)的倹{�?br>
查询:
假设要查询从m到n�q�一�D늚�最��? 那么我们先求��Z��个最大的k, 使得k满��2^k <= (n - m + 1).
于是我们��可以把[m, n]分成两个(部分重叠�?长度�?^k的区�? [m, m+2^k-1], [n-2^k+1, n];
而我们之前已�l�求��Z��f(m, k)为[m, m+2^k-1]的最��? f(n-2^k+1, k)为[n-2^k+1, n]的最��?br>我们只要�q�回其中更小的那�? ��是我们惌��的答�? �q�个��法的时间复杂度是O(1)�?
例如, rmq(0, 11) = min(f(0, 3), f(4, 3))
*/

#include<iostream>
#include<cmath>
using namespace std;
#define MAXN 1000000
#define mmin(a, b)   ((a)<=(b)?(a):(b))
#define mmax(a, b)   ((a)>=(b)?(a):(b))

int num[MAXN];
int f1[MAXN][100];
int f2[MAXN][100];

//��试输出所有的f(i, j)
void dump(int n)
{
    int i, j;
    for(i = 0; i < n; i++)
    {
        for(j = 0; i + (1<<j) - 1 < n; j++)
        {
            printf("f[%d, %d] = %d\t", i, j, f1[i][j]);
        }
        printf("\n");
    }
    for(i = 0; i < n; i++)
       printf("%d ", num[i]);
    printf("\n");
    for(i = 0; i < n; i++)
    {
        for(j = 0; i + (1<<j) - 1 < n; j++)
        {
            printf("f[%d, %d] = %d\t", i, j, f2[i][j]);
        }
        printf("\n");
    }
    for(i = 0; i < n; i++)
        printf("%d ", num[i]);
    printf("\n");
}

//sparse table��法
void st(int n)
{
    int i, j, k, m;
    k = (int) (log((double)n) / log(2.0));
    for(i = 0; i < n; i++)
    {
        f1[i][0] = num[i]; //递推的初�?/span>
        f2[i][0] = num[i];
    }
    for(j = 1; j <= k; j++)
    { //自底向上递推
        for(i = 0; i + (1 << j) - 1 < n; i++)
        {
            m = i + (1 << (j - 1)); //求出中间的那个�?/span>
            f1[i][j] = mmax(f1[i][j-1], f1[m][j-1]);
            f2[i][j] = mmin(f2[i][j-1], f2[m][j-1]);
        }
    }
}

//查询i和j之间的最�?注意i是从0开始的
void rmq(int i, int j)
{
    int k = (int)(log(double(j-i+1)) / log(2.0)), t1, t2; //用对2��d��数的�Ҏ��求出k
    t1 = mmax(f1[i][k], f1[j - (1<<k) + 1][k]);
    t2 = mmin(f2[i][k], f2[j - (1<<k) + 1][k]);
    printf("%d\n",t1 - t2);
}

int main()
{
    int i,N,Q,A,B;
    scanf("%d %d", &N, &Q);
    for (i = 0; i < N; ++i)
    {
        scanf("%d", num+i);
    }

    st(N); //初始�?br>    //dump(N); //��试输出所有f(i, j)
    while(Q--)
    {
        scanf("%d %d",&A,&B);
        rmq(A-1, B-1);
    }
    return 0;
}