hdu 3549 Flow Problem（最大流模板题）

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本文详细介绍了网络流中的最大流问题，并通过一个具体的加权有向图实例演示了如何求解最大流。提供了完整的C++代码实现，帮助读者理解和掌握最大流算法。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3549

Problem Description

Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.

Input

The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)

Output

For each test cases, you should output the maximum flow from source 1 to sink N.

Sample Input

    2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1
   

Sample Output

    Case 1: 1 Case 2: 2
   

代码例如以下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 32;//点数的最大值
const int MAXM = 1017;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
	int to, cap, flow;
	int next;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int dep[MAXN],pre[MAXN],cur[MAXN];
int gap[MAXN];//gap[x]=y :说明残留网络中dep[i]==x的个数为y
void init()
{
	tol = 0;
	memset(head,-1,sizeof (head));
}
//加边，单向图三个參数。双向图四个參数
void addedge (int u,int v,int w,int rw=0)
{
	edge[tol].to = v;edge[tol].cap = w;
	edge[tol].next = head[u];
	edge[tol].flow = 0;
	head[u] = tol++;
	edge[tol].to = u;edge[tol].cap = rw;
	edge[tol]. next = head[v];
	edge[tol].flow = 0;head[v]=tol++;
}
//输入參数：起点、终点、点的总数
//点的编号没有影响，仅仅要输入点的总数
int sap(int start,int end, int N)
{
	memset(gap,0,sizeof(gap));
	memset(dep,0,sizeof(dep));
	memcpy(cur,head,sizeof(head));
	int u = start;
	pre[u] = -1;
	gap[0] = N;
	int ans = 0;
	int i;
	while(dep[start] < N)
	{
		if(u == end)
		{
			int Min = INF;
			for( i = pre[u];i != -1; i = pre[edge[i^1]. to])
			{
				if(Min > edge[i].cap - edge[i]. flow)
					Min = edge[i].cap - edge[i].flow;
			}
			for( i = pre[u];i != -1; i = pre[edge[i^1]. to])
			{
				edge[i].flow += Min;
				edge[i^1].flow -= Min;
			}
			u = start;
			ans += Min;
			continue;
		}
		bool flag =  false;
		int v;
		for( i = cur[u]; i != -1;i = edge[i].next)
		{
			v = edge[i]. to;
			if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
			{
				flag =  true;
				cur[u] = pre[v] = i;
				break;
			}
		}
		if(flag)
		{
			u = v;
			continue;
		}
		int Min = N;
		for( i = head[u]; i !=  -1; i = edge[i]. next)
		{
			if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
			{
				Min = dep[edge[i].to];
				cur[u] = i;
			}
		}
		gap[dep[u]]--;
		if(!gap[dep[u]]) 
			return ans;
		dep[u] = Min+1;
		gap[dep[u]]++;
		if(u != start) 
			u = edge[pre[u]^1].to;
	}
	return ans;
}
int main()
{
	int n, m;
	int a, b, w;
	int c, s, t;
	int i;
	int T;
	int cas = 0;
	scanf("%d",&T);
	while(T--)
	{
		init();//初始化    
		scanf("%d%d",&n,&m);
		for(i = 1; i <= m; i++)//边数
		{
			scanf("%d%d%d",&a,&b,&w);
			addedge(a,b,w,0);
		//	addedge(b,a,w,0);
		}
		int ans = sap(1, n, n);
		printf("Case %d: %d\n",++cas,ans);
	}
	return 0;
}