B - Flow Problem

最新推荐文章于 2018-12-28 12:23:05 发布

Jack_zhuiyi

最新推荐文章于 2018-12-28 12:23:05 发布

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分类专栏：最大流

本文链接：https://blog.youkuaiyun.com/Jack_zhuiyi/article/details/79431424

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本文介绍了一种解决ACM竞赛中经典网络流问题的方法，通过构建加权有向图并利用深度优先搜索来寻找从源点到汇点的最大流量路径。文章提供了完整的C++实现代码，包括边的定义、添加边的函数、深度优先搜索算法及最大流计算。

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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.

InputThe 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)OutputFor each test cases, you should output the maximum flow from source 1 to sink N.Sample Input

Sample Output

Case 1: 1
Case 2: 2

套个模板A，暂时熟悉一下模板。

#include <iostream>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <cmath>
#include <stack>
#include <string>
#include <queue>
#include <algorithm>
#include <cstdio>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
struct edge {
	int cap;
	int to;
	int rev;
};
vector<edge> G[15];
bool flag[15];
void addedge(int x,int y,int cap) {
	G[x].push_back(edge{ cap,y,(int)G[y].size() });
	G[y].push_back(edge{0,x,(int)G[x].size() - 1});
}
int dfs(int k,int t,int f) {
	if (k == t) {
		return f;
	}
	int s = G[k].size();
	flag[k] = true;
	for (int i = 0; i < s; ++i) {
		if (!flag[G[k][i].to] && G[k][i].cap > 0) {
			int d = dfs(G[k][i].to, t, min(f, G[k][i].cap));
			if (d > 0) {
				G[k][i].cap -= d;
				G[G[k][i].to][G[k][i].rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
int ful(int s, int t){
	int flow = 0;
	while (1) {
		memset(flag, false, sizeof(flag));
		int f = dfs(s,t,INF);
		if (f == 0) {
			return flow;
		}
		flow += f;
	}
	return flow;
}
int main(){
	int n, m, T, t = 0;
	scanf("%d", &T);
	while (T--){
		t++;
		scanf("%d%d", &n, &m);
		int x, y, cap;
		for (int i = 1; i <= m; ++i) {
			scanf("%d%d%d", &x, &y,&cap);
			addedge(x, y, cap);
		}
		printf("Case %d: %d\n", t,ful(1,n));
		for (int i = 1; i <= n; ++i) {
			G[i].clear();
		}
	}
	return 0;
}