最大流入门——POJ 1273

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Drainage Ditches

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

Submit Status

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

Source

题意：求从源点到汇点的最大流，注意重边。

思路：不断用spfa找1~n的最短路，然后min=（最短路中距离最小的边长），ans+=min。再把最短路中每条边长都减去min，同时反向弧加上min，直到不存在增广路。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
const int MAXN=200+10;
#define ms(x,y) memset(x,y,sizeof(x))
#define INF 1<<30
#define LL long long
using namespace std;
int mark[MAXN];
int fa[MAXN];
int a[MAXN];
int cap[MAXN][MAXN];
int dis[MAXN];
int ans;

void bfs(int s, int t)
{
	for(;;){
		for(int i=1; i<=t; i++){
			dis[i] = INF;
			a[i]=INF;
		}
		ms(fa,0);
		ms(mark,0);
		dis[1] = 0;
		queue<int>q;
		q.push(s);
		mark[s]=1;
		while(!q.empty())
		{
			int u=q.front();
			q.pop();
			mark[u]=0;
			for(int v=1; v<=t; v++){
				if(cap[u][v] && dis[u] + cap[u][v] < dis[v]){
					dis[v] = dis[u] + cap[u][v];
					fa[v]=u;
					a[v] = min(a[u], cap[u][v]);
					if(mark[v]==0){
						q.push(v);
						mark[v]=1;
					}
				}
			}
		}
		if(dis[t] == INF) break;
		ans+=a[t];
		for(int id=t; id!=s; id=fa[id]){
			cap[fa[id]][id]-=a[t];
			cap[id][fa[id]]+=a[t];
		}
	}
}

int main()
{
	//freopen("in.txt","r",stdin);
	int n,m;
	while(~scanf("%d%d" ,&m,&n))
	{
		ans=0;
		ms(cap,0);
		int u,v,c;
		for(int i=0; i<m; i++){
			scanf("%d%d%d", &u,&v,&c);
			cap[u][v] += c;
		}
		bfs(1,n);
		printf("%d\n", ans);
	}
	return 0;
}