poj1273

                                                                                              Drainage Ditches

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 49349		Accepted: 18689

Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch. 
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network. 
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle. 

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

Sample Output

50

#include <cstdio>
#include <cstring>
#define MAXN 210

struct Matrix
{
    int c, f;    //容量，流量
};
Matrix Edge[MAXN][MAXN]; //流及容量(邻接矩阵)
int M, N;    //读入的排水沟(即弧)数目，和汇合结点（即顶点）数目
int s, t;    //源点(结点1)、汇点(结点N)
int residual[MAXN][MAXN];    //残留网络
int qu[MAXN*MAXN], qs, qe;    //队列、队列头和尾
int pre[MAXN];    //pre[i]为增广路上顶点i前面的顶点序号
int vis[MAXN];    //BFS算法中各顶点的访问标志
int maxflow, min_augment;    //最大流流量、每次增广时的可改进量

void find_augment_path( )    //BFS求增广路
{
    int i, cu;    //cu为队列头顶点
    memset( vis, 0, sizeof(vis) );
    qs = 0;  qu[qs] = s;    //s入队列
    pre[s] = s;  vis[s] = 1;  qe = 1;
    memset( residual, 0, sizeof(residual) );  memset( pre, 0, sizeof(pre) );
    while( qs<qe && pre[t]==0 )
    {
        cu = qu[qs];
        for( i=1; i<=N; i++ )
        {
            if( vis[i]==0 )
            {
                if( Edge[cu][i].c - Edge[cu][i].f >0 )
                {
                    residual[cu][i] = Edge[cu][i].c - Edge[cu][i].f;
                    pre[i] = cu;  qu[qe++] = i;  vis[i] = 1;
                }
                else if( Edge[i][cu].f>0 )
                {
                    residual[cu][i] = Edge[i][cu].f;
                    pre[i] = cu;  qu[qe++] = i;  vis[i] = 1;
                }
            }
        }
        qs++;
    }
}

void augment_flow( )    //计算可改进量
{
    int i = t, j;    //t为汇点
    if( pre[i]==0 )
    {
        min_augment = 0;  return;
    }
    j = 0x7fffffff;
    while( i!=s )    //计算增广路上可改进量的最小值
    {
        if( residual[pre[i]][i]<j )  j = residual[pre[i]][i];
        i = pre[i];
    }
    min_augment = j;
}

void update_flow( )    //调整流量
{
    int i = t;    //t为汇点
    if( pre[i]==0 )  return;
    while( i!=s )
    {
        if( Edge[pre[i]][i].c - Edge[pre[i]][i].f > 0 )
            Edge[pre[i]][i].f += min_augment;
        else if( Edge[i][pre[i]].f >0 )
            Edge[pre[i]][i].f+=min_augment;
        i = pre[i];
    }
}

void solve( )
{
    s = 1;  t = N;
    maxflow = 0;
    while( 1 )
    {
        find_augment_path( ); //BFS寻找增广路
        augment_flow( );    //计算可改进量
        maxflow += min_augment; //
        if( min_augment>0 )  update_flow( ); //更新流
        else return;
    }
}

int main( )
{
    int i;
    int u, v, c;
    while( scanf("%d %d",&M,&N)!=EOF )
    {
        memset( Edge, 0, sizeof(Edge) );
        for( i=0; i<M; i++ )
        {
            scanf( "%d %d %d", &u, &v, &c );
            Edge[u][v].c += c;
        }
        solve( );
        printf( "%d
", maxflow );
    }
    return 0;
}