poj 1273 Drainage Ditches

<span style="font-size: 18pt; font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">Drainage Ditches</span>

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 54939		Accepted: 20946

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

给出关键部分伪代码来介绍一下网络流的标号法：
Repeat
    队列置空；
    全部点设为未标号；
    将源点增加队列，并标号为（0，+∞）；
    while 队列非空
        {
            头指针+1
            依次检查与头指针指向的元素相连的边
            if 还有一点没有标号 and 流量可改进
                {
                    尾指针+1,该点入队
                    a[还有一点]<-队列头指针元素
                    b[还有一点]<-Min{b[头指针元素],边的最大流量};
                }
        }
        if 汇点已标号
            {
                从汇点出发依次改动各条边的流量
            }
Until 汇点未标号；
答案<-全部与汇点相连的边的流量

EK算法：是一种最短路径增值的算法。通过不断从源点广搜寻找最短路径，然后记录路径中的最小容量。再给这条路径上的边上flow增值，（增值之后当然会有一部分边是满流的，那么再次广搜的时候当然也就不能正向搜索到此边了，这条路径上的边的流量都增大了，容量不变，可增值量当然也就会降低），直到从源点广搜不到汇点为止，来实现最大流。
因为每次都要广搜所以时间复杂度会达到O（m*m+n）,m为边的个数。n为点的个数。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<string.h>
#include<queue>
#include<limits.h>
#define MAX 250
using namespace std;
int cap[MAX][MAX];
int m;
int EKarp(int s,int t)
{
	queue<int> Q;
	int flow[MAX][MAX],a[MAX],u,v,f,pre[MAX];
	f=0;
	memset(flow,0,sizeof(flow));
	while(1)
	{
		Q.push(s);
		memset(a,0,sizeof(a));
		a[s]=INT_MAX;
		while(!Q.empty())
		{
			u=Q.front();
			Q.pop();
			for(v=1;v<=m;v++)
				if(!a[v]&&cap[u][v]>flow[u][v])
				{
					
					Q.push(v);
					a[v]=a[u]<cap[u][v]-flow[u][v]?a[u]:cap[u][v]-flow[u][v];
					pre[v]=u;
				}
		}
		if(a[t]==0)break;
		for(u=t;u!=s;u=pre[u])
		{
			flow[pre[u]][u]+=a[t];
			flow[u][pre[u]]-=a[t];
		}
		f+=a[t];
	}
	return f;
}
int main()
{
	int from,to,c,n,ans;
	while(~scanf("%d%d",&n,&m))
	{
		memset(cap,0,sizeof(cap));
		while(n--)
		{
			scanf("%d%d%d",&from,&to,&c);
			cap[from][to]+=c;
		}
		ans=EKarp(1,m);
		printf("%d
",ans);
	}
	return 0;
}