hdoj 3549 Flow Problem 【最大流入门 dinic算法】

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 9735    Accepted Submission(s): 4562

 

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

第一道最大流，dinic算法：

#include 
#include 
#include 
#include 
#include 
#define MAXN 100+10
#define MAXM 2000+10
#define INF 10000000+10
using namespace std;
struct Edge
{
	int from, to, cap, flow, next;
}edge[MAXM];
int n, m;
int dist[MAXN];//距离起点的距离      
int top, head[MAXN], vis[MAXN];
int cur[MAXN];//cur[i]表示i点正在考虑的边 优化不再考虑已经用过的点 初始化为head 
void init()
{
	top = 0;
	memset(head, -1, sizeof(head));
}
void addedge(int a, int b, int c)
{
	Edge E1 = {a, b, c, 0, head[a]};
	edge[top] = E1;
	head[a] = top++;
	Edge E2 = {b, a, 0, 0, head[b]};
	edge[top] = E2;
	head[b] = top++;
}
void getmap()
{
	int a, b, c;
	while(m--)
	{
		scanf("%d%d%d", &a, &b, &c);
		addedge(a, b, c);
	}
}
bool BFS(int start, int end)
{
//从残余网络找一条从start->end的路径(就是最短路的代码,每条边的边权为最大流量-当前流量,当这条边满流时最大流量=当前流量，边权为0）   
//显然当找不到这样的路径时说明不能继续增加流量 则返回false 算法结束   	
	int i;
	memset(dist, -1, sizeof(dist));
	memset(vis, 0, sizeof(vis));
	queue Q;
	while(!Q.empty()) Q.pop();
	Q.push(start);
	vis[start] = 1;
	dist[start] = 0;
	while(!Q.empty())
	{
		int u = Q.front();
		Q.pop();
		for(i = head[u]; i != -1; i = edge[i].next)
		{
			Edge E = edge[i];
			if(!vis[E.to] && E.cap > E.flow)
			{
				vis[E.to] = 1;
				dist[E.to] = dist[u] + 1;
				if(E.to == end) return true;
				Q.push(E.to); 
			}
		}
	}
	return false;
}
int DFS(int x, int a, int end)//增广路径 叠加当前start->end路径上的最小残量 a 
{
	if(x == end || a == 0) return a;
	int flow = 0, f;
	for(int& i = cur[x]; i != -1; i = edge[i].next)//从上次考虑的弧 
	{
		Edge& E = edge[i];
		if(dist[x]+1 == dist[E.to] && (f = DFS(E.to, min(a, E.cap-E.flow), end)) > 0)
		{
			E.flow += f;
			edge[i^1].flow -= f;
			flow += f;
			a -= f;
			if(a == 0) break;
		}
	}
	return flow;
}
int Maxflow(int start, int end)
{
	int flow = 0;
	while(BFS(start, end))//找到1到n的最短路径 
	{
		memcpy(cur, head, sizeof(head));
		flow += DFS(start, INF, end);
	}
	return flow;
}
int main()
{
	int t;
	int k = 1;
	scanf("%d", &t);
	while(t--)
	{
		scanf("%d%d", &n, &m);
		init();
		getmap();
		printf("Case %d: %d\n", k++, Maxflow(1, n));
	}
	return 0;
}