Drainage Ditches

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

这道题最开始以为是什么最小生成树，后来看了书才明白这是最大流的问题，可以说这是我的第一道最大流，我在图论书271找到了和他很类似的问题，用最短增广路就可以解决。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int map[210][210],book[210],s[210],st,en,n,m;
int bfs()
{
    int i,t;
    queue<int>q;
    memset(s,-1,sizeof(s));
    book[st]=INF;
    s[st]=1;
    while(!q.empty())
    {
        q.pop();
    }
    q.push(st);
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        if(t==en)
        {
            break;
        }
        for(i=1;i<=m;i++)
        {
            if(i!=st && s[i]==-1 && map[t][i])
            {
                s[i]=t;
                if(book[t]< map[t][i])
                    book[i]=book[t];
                else
                    book[i]=map[t][i];
                q.push(i);
            }
        }
    }
    if(s[en]==-1)
    {
        return -1;
    }
    else
    {
        return book[en];
    }
}
int cn()
{
    int flag=0,ans,x,y;
    while((ans=bfs())!=-1)
    {
        flag+=ans;
        y=en;
        while(y!=st)
        {
            x=s[y];
            map[x][y]-=ans;
            map[y][x]+=ans;
            y=x;
        }
    }
    return flag;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int a,b,l;
        memset(map,0,sizeof(map));
        for(int i=0;i<n;i++)
        {
            scanf("%d%d%d",&a,&b,&l);
            map[a][b]+=l;
        }
        en=m;
        st=1;
        printf("%d\n",cn());
    }
    return 0;
}