Drainage Ditches 【HDU - 1532】

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

题意：

这是一个网络流问题。

N是农夫约翰挖的沟渠的数量。M是这些沟渠交叉点的数量。

下列N行中的每一行都包含三个整数，Si、Ei和Ci。Si和Ei (1 <= Si, Ei <= M)指定这条沟渠流动的交叉点。水将通过这条沟渠从Si流到Ei。Ci (0 <= Ci <= 10,000,000)是水流通过沟渠的最大流速。

对于每种情况，输出一个整数，即从池中流出的水的最大速率。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;

#define inf 0x3f3f3f3f

int n,m;
int a,b,c;
int s[210],k[210],mapp[210][210];

int bfs()
{
    memset(k,-1,sizeof(k));
    queue<int>q;
    k[1] = 0;
    s[1] = inf;
    q.push(1);
    while(!q.empty())
    {
        int i;
        int now = q.front();
        q.pop();
        for(i = 1; i <= m; i ++ )
        {
            if( i != 1 && mapp[now][i] && k[i]== -1)
            {
                k[i] = now;
                s[i] = min(mapp[now][i],s[now]);
                q.push(i);
            }
        }
    }
    if(k[m]==-1)
        return -1;
    return s[m];
}

int dfs()
{
    int sum=0,d,e,f;
    while(1)
    {
        d=bfs();
        if(d == -1)
            break;
        sum += d;
        e = m;
        while(e != 1)
        {
            f = k[e];
            mapp[f][e] -= d;
            mapp[e][f] += d;
            e = f;
        }
    }
    return sum;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(mapp,0,sizeof(mapp));
        for(int i = 1; i <= n; i ++)
        {
            scanf("%d%d%d",&a,&b,&c);
            mapp[a][b]+=c;
        }
        printf("%d\n",dfs());
    }
    return 0;
}