HDU 1532 Drainage Ditches 分类： Brush Mode 2014-07-31 10:38 82人阅读 评论(0) 收藏

Drainage Ditches

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8861    Accepted Submission(s): 4141

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

 

题意不说了，这是网络流经典入门题，最大的坑就是，在最短路题目里面，重边取最小，在网络流里面，重边就相加

WA了N次，才发现这个严重的问题

#include<cstdio>
#include<cstring>
#include<limits.h>
#define min(a,b)	a>b?b:a
using namespace std;
int mi[35][5];
int vis[10005];
int m;
int g[1005][1005];
int dfs(int si,int ti,int f)
{
	if(si==ti)	return f;
	vis[si] = 1;
	for(int i=1;i<=m;i++)
	{
		if(!vis[i] && g[si][i]>0)
		{
			int d = dfs(i,ti,min(f,g[si][i]));
			if(d)
			{
				g[si][i]-=d;
				g[i][si]+=d;
				return d;
			}
		}
	}
	return 0;
}

int find(int s, int t)
{
	int res = 0;
	for(;;)
	{
		memset(vis,0,sizeof(vis));
		int f;
		f = dfs(s,t,INT_MAX);
		if(f==0)	return res;
		res+=f;
	}
}
int main()
{
	int a, b, c;
	int n;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(g,0,sizeof(g));
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			g[a][b] += c;
		}
		printf("%d\n",find(1,m));
	}
	return 0;
}   

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