Drainage Ditches
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16292 Accepted Submission(s): 7737
Problem 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
题意
有N个沟，M个点，有个河Stream(源点) ，有个池塘Pond(汇点)，求这个汇点最多能存多少水 。
分析
和之前的hdu3549-Flow Problem 基本一样，熟悉的题意，类似的坑，只是这会我在网上找了个好的模板,比我代码风格简洁多了, %%%%%%ORZ
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn=1000;
int n,m;
int s,t;
int vis[maxn];
int pre[maxn];
int cap[maxn][maxn];
int BFS(){
queue<int> q;
memset(pre,-1,sizeof(pre));
memset(vis,0,sizeof(vis));
pre[s]=s;
vis[s]=1;
q.push(s);
while(! q.empty()){
int cur= q.front();
q.pop();
for(int i=1;i<=n;i++){
if( !vis[i] && cap[cur][i] >0){
pre[i]=cur;
vis[i]=1;
if( i== t)
return 1;
q.push(i);
}
}
}
return false;
}
int edmonds_karp(){
int ans=0;
while(BFS()){
int minn= INF;
for(int i=t; i != s; i=pre[i])
minn= min(minn, cap[pre[i]][i]);
for(int i=t;i !=s ; i=pre[i]){
cap[pre[i]][i] -=minn;
cap[i][pre[i]] +=minn;
}
ans +=minn;
}
return ans;
}
int main(){
//freopen("in.txt","r",stdin);
while(cin>>m>>n){
memset(cap,0,sizeof(cap));
int t1,t2,t3;
for(int i=0;i<m;i++){
cin>>t1>>t2>>t3;
cap[t1][t2] +=t3;
}
s=1;
t=n;
cout<<edmonds_karp()<<endl;
}
return 0;
}