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：给出n个河流（挖的沟），m个点（沟与沟的交叉点），以及每个河流的流量，求从1到m点的最大流量。1是池塘，m是小溪。
题解：最大流模板题，详细见代码。
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MAX = 500;
int n,m;
int map[MAX][MAX],flow[MAX][MAX];
int p[MAX],a[MAX];
int ekmaxflow(int s,int t){
int sum=0;
int i;
queue<int>q;
memset(flow,0,sizeof(flow));
while(true){//没有其他增广路线时停止
memset(a,0,sizeof(a));//记住残余水量
a[s]=inf;
q.push(s);
while(!q.empty()){
int u=q.front();
q.pop();
for(i=2;i<=m;i++){
if(!a[i]&&map[u][i]>flow[u][i]){
p[i]=u;
q.push(i);
a[i]=a[u]<map[u][i]-flow[u][i]?a[u]:map[u][i]-flow[u][i];
}
}
}
if(!a[t])////为 0 时离开
break;
for(i=t;i!=s;i=p[i]){
flow[p[i]][i]+=a[t];
flow[i][p[i]]-=a[t];
}
sum=sum+a[t];
}
return sum;
}
int main(){
while(cin >> n >> m){
memset(map,0,sizeof(map));
memset(p,0,sizeof(p));//记住每个点的父亲是哪儿位
while(n--){
int a,b,c;
cin >> a >> b >> c;
map[a][b]=map[a][b]+c;//存两个点之间的流量
}
int maxx=ekmaxflow(1,m);
cout << maxx <<endl;
}
return 0;
}小菜鸡的我不是很懂，不懂样例为什么是50，讲真，板子挺猛，这题，娘的，呵呵呵~~~