国产一区二区三区不卡在线观看,久久免费精品视频,欧美一级视频精品观看

AOJ 236 Cow Picnic , poj 3256

田兵 — Mon, 30 Aug 2010 07:49:00 GMT

Cow Picnic

Time Limit: 2000 ms Memory Limit: 64 MB
Total Submission: 1 Accepted: 1

Description

The cows are having a picnic! Each of Farmer John's K (1 ≤ K ≤ 100) cows is grazing in one of N (1 ≤ N ≤ 1,000) pastures, conveniently numbered 1...N. The pastures are connected by M (1 ≤ M ≤ 10,000) one-way paths (no path connects a pasture to itself).

The cows want to gather in the same pasture for their picnic, but (because of the one-way paths) some cows may only be able to get to some pastures. Help the cows out by figuring out how many pastures are reachable by all cows, and hence are possible picnic locations.

Input

Line 1: Three space-separated integers, respectively: K, N, and M
Lines 2..K+1: Line i+1 contains a single integer (1..N) which is the number of the pasture in which cow i is grazing.
Lines K+2..M+K+1: Each line contains two space-separated integers, respectively A and B (both 1..N and A != B), representing a one-way path from pasture A to pasture B.

Output

Line 1: The single integer that is the number of pastures that are reachable by all cows via the one-way paths.

Sample Input

2 4 4
2
3
1 2
1 4
2 3
3 4

Sample Output

2[EOL][EOF]

Hint

The cows can meet in pastures 3 or 4.

Source

USACO 2006 December Silver

从每个牛开始求一�ơ单源最短�\径，假设��L(f��ng)��是X�Q�如果从X能到i (di[i]!=INF) �Q�cnt[i]++�Q�用来统计能到达 i 点的牛的数量�?br>
�l�果��是满��cnt[i]==K的数量，即i�Ҏ(gu��)��有的牛都可以到达�?br>
用spfa求，spfa在这里不是求最�D��\径，只要��C��(ji��n)��p��Q�不需要是最短的�Q�因此会(x��)更快一炏V�?br>

#include<iostream>
#include<time.h>
#include<vector>
#include<queue>
using namespace std;
const int MAX=1001,INF=0x0fffffff;
vector<int> mp[MAX];
int d[MAX], cnt[MAX];
int K,N,M;
int stay[101];
void spfa(int x)
{
     for(int i=1; i<=N; i++)
             d[i]=INF;
     queue<int>q;
     q.push(x);
     d[x]=0;
     while(q.size())
     {  
         
          int u=q.front(); q.pop(); 
          for(int i=0; i<mp[u].size(); i++)
          {
                  if(d[mp[u][i]]==INF)
                  {
                       d[mp[u][i]]=d[u]+1;
                       q.push(mp[u][i]);
                  }
          }
                    
     }
}

int main()
{
    cin>>K>>N>>M;
    
    for(int i=1; i<=K; i++)
            cin>>stay[i];
    
    for(int i=1,s,t; i<=M; i++)
            {
                     cin>>s>>t;
                     mp[s].push_back(t);
            }
    
    for(int i=1; i<=K; i++)
    {
       spfa(stay[i]);    
       for(int i=1; i<=N; i++)
               if(d[i]!=INF)cnt[i]++;   
    }
    
    int ans=0;
    
    for(int i=1; i<=N; i++)  
            if(cnt[i]==K)ans++;
    
    cout<<ans<<endl;        
    system("pause");
    return 0;
}

田兵 2010-08-30 15:49 发表评论

poj 1274 The Perfect Stall

田兵 — Mon, 30 Aug 2010 02:09:00 GMT

二分囄��最大匹配，加一个源点，�q�个源点到左�Ҏ(gu��)��有顶点加一条边�Q�权��gؓ(f��)1. 。同理，加一个终点，所有右边的定点到这个终�Ҏ(gu��)��条边�Q�权��gؓ(f��)1. �q�样二分囄��最大匹配就转换成了(ji��n)最大流问题�?/span>

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
const int MAX=405;
int cap[MAX][MAX]={0};
int flow[MAX][MAX]={0};
int pre[MAX],m[MAX];
int N, M,S,T;
const int INF=10000000;

int bfs(int s)
{
    memset(m,0,sizeof m);  
    memset(pre,0,sizeof pre);
    queue<int> q;
    q.push(s);
    m[s]=INF;
    while(!q.empty())
    {
            int u=q.front(); q.pop();
            for(int v=0; v<=N+M+1; v++)
                    if(!m[v]&&cap[u][v]>flow[u][v])
                    {
                         pre[v]=u;
                         m[v]= m[u]>cap[u][v]-flow[u][v]?cap[u][v]-flow[u][v]:m[u];
                         q.push(v);
                    }
    }
    
   if(m[T]==0)return 0;
   
   for(int u=T; u!=S ; u=pre[u])
   {
           flow[pre[u]][u]+=m[T];
           flow[u][pre[u]]-=m[T];
   }
   return m[T];
}

int main()
{
    
    while(cin>>N>>M)
    {    memset(cap,0,sizeof cap);
         memset(flow,0,sizeof flow);
         for(int i=1; i<=N; i++)
         {
            int c,e;
            cin>>c;
            for(int j=1; j<=c; j++)
                   {
                         cin>>e;
                         e=e+N;
                         cap[i][e]=1;
                   } 
         } 
         S=0; 
         T=N+M+1;
         for(int i=1; i<=N; i++)
            cap[0][i]=1;
            
         for(int i=1; i<=M; i++)
            cap[N+i][T]=1;
    
         int ans=0;
         while(1)
         {
            int temp=bfs(S);
            if(temp==0)break;
            else ans+=temp;
         }
    
         cout<<ans<<endl;
    }   
    system("pause");
    return 0;
}

田兵 2010-08-30 10:09 发表评论

poj 1273 Drainage Ditches

田兵 — Wed, 25 Aug 2010 07:30:00 GMT

         赤裸裸的最大流问题�Q�也是我写第一个最大流�Q�是入门的好题�?br>
         发现看代码有时候比看书更容易懂�Q�看最大流已经看了(ji��n)好几天书�?ji��n)，都是晕乎乎的�Q�参考了(ji��n)一下别人的代码�Q�似乎有些了(ji��n)解了(ji��n)�?br>
      Ford-Fulkerson�Ҏ(gu��)��的增�q��\��法
      每次用dfs��L��可增�q��\径，该�\径上的所有边的容量均要是正数�Q�如果找到就取该路径上所有容量的最��g��为增加的��。如果不存在则说明已�l�是最大流�Q�结束�?br>
下面�q�两个程序在hdu有点��问题，�W�三个是在两个OJ上都可以的�?br>

#include<iostream>
using namespace std;
int map[250][250];
int pre[250];
bool visit[250];
int m,n,minflow;
const int INF=0x7fffffff/1000;

int dfs(int v)
{
    visit[v]=true;
    if(v==n)return 1;
    for(int i=1; i<=n; i++)
            if(map[v][i]>0&&!visit[i])
              { 
                  pre[i]=v;
                  minflow=minflow>map[v][i]?map[v][i]:minflow;
                  if(dfs(i))return true;
              }
    return 0;
}

int maxflow()
{
    int maxf=0;
    while(1)
    {
           memset(pre,0,sizeof pre);
           minflow=INF;
           memset(visit,0,sizeof visit);
           pre[1]=-1;
           if(dfs(1))
           {
                    
                     int inc=minflow;
                     int t=n;
                     while(pre[t]!=-1)
                     {
                          int qq=pre[t];
                          map[qq][t]-=inc;
                          map[t][qq]+=inc;
                          t=qq;
                     }
                     maxf+=inc;
           }
           else break;
    }
    return maxf;
}

int main()
{
    while(cin>>m>>n)
    {
       int s,t,c;
       memset(map,0,sizeof map);
       while(m--)
       {
                cin>>s>>t>>c;
                map[s][t]+=c; 
       }               
       cout<<maxflow()<<endl;
    }
    
   system("pause");
   return 0;   
}

用BFS的Edmonds-Karp��法

#include<iostream>
#include<queue>
using namespace std;
int map[250][250];
int pre[250];
bool visit[250];
int m,n,minflow;
const int INF=0x7fffffff/1000;

int bfs(int s)
{
    queue<int> Q;
    Q.push(s);
    visit[s]=true;
    while(Q.size())
    {
         int v=Q.front(); Q.pop();
         if(v==n)return true ; 
         for(int i=1; i<=n; i++)
                 if(!visit[i]&&map[v][i]>0&&!visit[n])
                 {  
                     pre[i]=v;
                     visit[i]=true;
                     minflow=minflow>map[v][i]?map[v][i]:minflow;
                     Q.push(i);
                 }
                   
    }
    return false;
}

int maxflow()
{
    int maxf=0;
    while(1)
    {
           memset(pre,0,sizeof pre);
           minflow=INF;
           memset(visit,0,sizeof visit);
           pre[1]=-1;
           if(bfs(1))
           {
                     int inc=minflow;
                     int t=n;
                     while(pre[t]!=-1)
                     {
                          int qq=pre[t];
                          map[qq][t]-=inc;
                          map[t][qq]+=inc;
                          t=qq;
                     }
                     maxf+=inc;
           }
           else break;
    }
    return maxf;
}

int main()
{
    while(cin>>m>>n)
    {
       int s,t,c;
       memset(map,0,sizeof map);
       while(m--)
       {
                cin>>s>>t>>c;
                map[s][t]+=c; 
       }               
       cout<<maxflow()<<endl;
    }
    
   system("pause");
   return 0;   
}

问题或许��出现在取最��值的问题上，不过我认为前两种每次取的可能不是最��的�Q�但应该是没错的�?br>最多多dfs或bfs几次�?br>

#include<iostream>
#include<string.h>
using namespace std;
int map[250][250];
int pre[250],q[250];
bool visit[250];
int m,n;
const int INF=0x7fffffff;

bool dfs(int v)
{
    visit[v]=true;
    if(v==n)return true;
    for(int i=1; i<=n; i++)
            if(map[v][i]>0&&!visit[i])
              { 
                  pre[i]=v;
                  q[i]=q[v]>map[v][i]?map[v][i]:q[v];
                  if(dfs(i))return true;
              }
    return 0;
}

int maxflow()
{
    int maxf=0;
    while(1)
    {
           memset(pre,0,sizeof pre);
           memset(q,0,sizeof q);
           //for(int i=1; i<=n; i++)
                   q[1]=INF;
           memset(visit,0,sizeof visit);
           pre[1]=-1;
           if(dfs(1))
           {
                    
                     int inc=q[n];
                     int t=n;
                     while(pre[t]!=-1)
                     {
                          int qq=pre[t];
                          map[qq][t]-=inc;
                          map[t][qq]+=inc;
                          t=qq;
                     }
                     maxf+=inc;
           }
           else break;
    }
    return maxf;
}

int main()
{
    while(cin>>m>>n)
    {
       int s,t,c;
       memset(map,0,sizeof map);
       while(m--)
       {
                cin>>s>>t>>c;
                map[s][t]+=c; 
       }               
       cout<<maxflow()<<endl;
    }
    
   system("pause");
   return 0;   
}

田兵 2010-08-25 15:30 发表评论

田兵 — Thu, 19 Aug 2010 01:38:00 GMT

题意�Q?br> ��是从各点到X的最短距��d��(qi��ng)从X到各点的最短距��d��的最大倹{�?br>
�W�一利用dijstra求出从X到各点的最短距��R�?br>
然后所有的边反向，再进行一�ơdijstra求X到各点的最短�\径�?br>
�W�二�ơ求出的最短�\径也��是各点到X的最短�\径，因�ؓ(f��)边已�l�反向，对于�W�二�ơ从X到各点的最短�\径正�?br>原图从各点到X的最短�\径�?br>

#include<iostream>
#include<queue>
using namespace std;
const int INF=0x7fffffff/100;
int map[1001][1001]={0};
int d[1001]={0},N,M,X;
int dd[1001]={0};
bool f[1001];
void dijstra()
{
     for(int i=1; i<=N; i++)
             d[i]=INF;
     d[X]=0;
     memset(f,0,sizeof f);
     for(int i=1; i<=N; i++)
     {
             int min=INF,u=0;
             for(int j=1; j<=N; j++)
                     if(d[j]<min&&!f[j])
                     {
                                 min=d[j];
                                 u=j;
                     }
             f[u]=true;
             for(int j=1; j<=N; j++)
             {
                     if(d[u]+map[u][j]<d[j])
                        d[j]=d[u]+map[u][j];
             }
     }
     
}

void traverse()
{
     int temp;
     for(int i=1; i<=N; i++)
     for(int j=1; j<i; j++)
     {
             temp = map[i][j];
             map[i][j]=map[j][i];
             map[j][i]=temp;
     }
}

int main()
{

    cin>>N>>M>>X;
    
    for(int i=1; i<=N; i++)
    for(int j=1; j<=N; j++)
            map[i][j]=INF;
    for(int i=1,s,t,value; i<=M; i++)
    {
          cin>>s>>t>>value;
          map[s][t]=value;
    }
    
    dijstra();
    for(int i=1; i<=N; i++)
             dd[i]=d[i]; 
            
    traverse();
    dijstra();
    int max=0;
    for(int i=1; i<=N; i++)
            if(d[i]+dd[i]>max)max=d[i]+dd[i];
    
    cout<<max<<endl;
    
    system("pause");
    return 0;
}

田兵 2010-08-19 09:38 发表评论

田兵 — Sat, 29 May 2010 14:03:00 GMT

The Max Weight

Time Limit: 1000 ms Memory Limit: 64 MB
Total Submission: 36 Accepted: 4

Description

There a lot of bridges connect different positions in Venice,but they can't carry too much weigh,so each of them has a limit which can be described as an interger.A man wants to carry some goods from positon 1 to n.Help him find how much can he carry.

Input

There are T cases.
For each case,the number of positions ( 1 < = n < = 100) and number m of bridges are exhibited on the first line.The following m lines contain triples of integers specifying start and end positions of the bridge and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one bridge between each pair of crossings.

Output

The output for every scenario begins with a line containing "Case #i:", where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that the man can transport. Terminate the output for the scenario with a blank line.

Sampel Input

1
3 3
1 2 3
1 3 4
2 3 5

Sample Output

Case #1:[EOL]
4[EOF]

题意�Q?br>n个点�Q�有些点间有桥，桥上有最大承重量�Q�问你从1到n可以最大携带的物品的重量�?br>题解�Q?nbsp;
Folyd��法变�Ş�Q�把求最短�\径的和改为求最大蝲重量的问题，关键是dis[i][j]=dis[i][j]>dis[i][k]+dis[k][j]?dis[i][j]>dis[i][k]+dis[k][j]?:dis[i][j];换成dis[i][j]=max(dis[i][j],min(dis[i][k],dis[k][j]));

 1#include<iostream>
 2#include<cmath>
 3#include<string.h>
 4using namespace std;
 5long long dis[105][105];
 6 
 8void Floyd(int n)
 9{
10     for(int k=1; k<=n; k++)
11     for(int i=1; i<=n; i++)
12     for(int j=1; j<=n; j++)
13     {
14       if(i!=k&&j!=k&&dis[i][k]&&dis[k][j])
15         dis[i][j]=max(dis[i][j],min(dis[i][k],dis[k][j]));
16     }
17}
18
19int main()
20{
21    int t,i,j,m,n;
22    cin>>t;
23    for(int k=1; k<=t; k++)
24    {
25      cin>>n>>m;
26      memset(dis,0,sizeof (dis));
27      i=1;
28      for(int s,e,w; i<=m; i++)
29      {
30        cin>>s>>e>>w;
31        dis[s][e]=dis[e][s]=w;
32      }
33      
34      Floyd(n);
35      
36     cout<<"Case #"<<k<<':'<<endl<<dis[1][n]<<endl<<endl;
37    }
38    return 0;
39}

田兵 2010-05-29 22:03 发表评论

USACO--AOJ Bessie Come Home --Floyd��法

田兵 — Sun, 23 May 2010 12:03:00 GMT

Bessie Come Home

Time Limit:JAVA/Others2000/1000MS Memory Limit:JAVA/Others131072/65536KB
Total Submit:6 Accepted:2

Description

It's dinner time, and the cows are out in their separate pastures. Farmer John rings the bell so they will start walking to the barn. Your job is to figure out which one cow gets to the barn first (the supplied test data will always have exactly one fastest cow).

Between milkings, each cow is located in her own pasture, though some pastures have no cows in them. Each pasture is connected by a path to one or more other pastures (potentially including itself). Sometimes, two (potentially self-same) pastures are connected by more than one path. One or more of the pastures has a path to the barn. Thus, all cows have a path to the barn and they always know the shortest path. Of course, cows can go either direction on a path and they all walk at the same speed.

The pastures are labeled `a'..`z' and `A'..`Y'. One cow is in each pasture labeled with a capital letter. No cow is in a pasture labeled with a lower case letter. The barn's label is `Z'; no cows are in the barn, though.

Input

Line 1:	Integer P (1 <= P <= 10000) the number of paths that interconnect the pastures (and the barn)
Line 2..P+1:	Space separated, two letters and an integer: the names of the interconnected pastures/barn and the distance between them (1 <= distance <= 1000)

Output

A single line containing two items: the capital letter name of the pasture of the cow that arrives first back at the barn, the length of the path followed by that cow.

Sample Input

5
A d 6
B d 3
C e 9
d Z 8
e Z 3

Sample Output

B 11

1Floyd ��法�Q?a >http://icpc.ahu.edu.cn:8080/AOJ/   做的�W�一个图论题
2囄��最短�\径问题，�?#8216;Z’的最短�\径；
3Floyd��法大概知道怎么用了(ji��n) �Q�好像是动态规划实现的�Q�不知道��Z��么这��h��对的
4O�Q�N^3�Q�求解最短�\径问题，数据范围��过400可能��危险了(ji��n)
5#include<iostream>
6#include<string.h>
7using namespace std;
8int dis[53][53];
9const int INF=10000000;
10void Floyd(int n)
11{
12     for(int k=1; k<=n; k++)
13     for(int i=1; i<=n; i++)
14     for(int j=1; j<=n; j++)
15      if(i!=k&&k!=j&&i!=j&&dis[i][k]+dis[k][j]<dis[i][j])
16      dis[i][j]=dis[i][k]+dis[k][j];
17
18}
19
20
21int main()
22{
23    int p,i,j,k,d,n1,n2;
24    cin>>p;
25    memset(dis,0,sizeof (dis));
26    for(i=1; i<=52; i++)
27    for(j=1; j<=52; j++)
28    dis[i][j]=INF;
29
30    for(i=1; i<=p; i++)
31    {
32       char v1,v2;
33       cin>>v1>>v2>>d;
34       if(v1==v2)continue;
35       n1=(v1>='a'?v1-'a'+1:v1-'A'+26+1);
36       n2=(v2>='a'?v2-'a'+1:v2-'A'+26+1);
37       if(d<dis[n1][n2])dis[n1][n2]=dis[n2][n1]=d;
38    }
39
40    Floyd(52);
41
42    int min=INF+100;
43    char c;
44    for(i=27; i<=51; i++) //大写字母到Z
45    {
46       if(dis[i][52]<min){min=dis[i][52];c=i; }
47    }
48    cout<<char(c-27+'A')<<' '<<min<<endl;
49    //system("pause");
50    return 0;
51}
52

田兵 2010-05-23 20:03 发表评论

国产一区二区三区不卡在线观看,久久免费精品视频,欧美一级视频精品观看

AOJ 236 Cow Picnic , poj 3256

poj 1274 The Perfect Stall

poj 1273 Drainage Ditches

USACO--AOJ Bessie Come Home --Floyd���法

USACO--AOJ Bessie Come Home --Floyd��法