POJ 2914 Minimum Cut 无向图最小割SW算法_minimum cut of an undirected graph-优快云博客

本文链接：https://blog.youkuaiyun.com/sinat_35406909/article/details/76794346

该博客介绍了如何求解无向图的最小割问题，具体是通过解决POJ 2914题目来阐述。内容包括输入输出格式说明，以及在最大生成树基础上应用堆优化的方法，实现时间复杂度为O(n^2logn)，适用于稀疏图。对于稠密图，通常采用O(n^3)的解决方案。

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Minimum Cut

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 9996		Accepted: 4177
Case Time Limit: 5000MS

Description

Given an undirected graph, in which two vertices can be connected by multiple edges, what is the size of the minimum cut of the graph? i.e. how many edges must be removed at least to disconnect the graph into two subgraphs?

Input

Input contains multiple test cases. Each test case starts with two integers N and M (2 ≤ N ≤ 500, 0 ≤ M ≤ N × (N − 1) ⁄ 2) in one line, where N is the number of vertices. Following are M lines, each line contains M integers A, B and C (0 ≤ A, B < N, A ≠ B, C > 0), meaning that there C edges connecting vertices A and B.

Output

There is only one line for each test case, which contains the size of the minimum cut of the graph. If the graph is disconnected, print 0.

Sample Input

Sample Output

2
1
2

Source

Baidu Star 2006 Semifinal
Wang, Ying (Originator)
Chen, Shixi (Test cases)

模板题。

复习网址：http://www.cnblogs.com/ylfdrib/archive/2010/08/17/1801784.html

在最大生成树中加入堆优化，复杂度可提升为O(n^2logn)。一般建议在稀疏图中加堆优化，稠密图就老老实实O(n^3)吧。

//无向图最小割SW Algorithm 
#include <cstdio>
#include <iostream>
#include <string.h>
#include <string> 
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <math.h>
#include <cmath>
#include <bitset>
#define mem0(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
using namespace std;
typedef long long ll;
typedef long double ld;
const int maxn=505,inf=0x3f3f3f3f;
const ll llinf=0x3f3f3f3f3f3f3f3f; 
const ld pi=acos(-1.0L);
int mp[maxn][maxn],dist[maxn];
int num,s,t,mcut;
bool visit[maxn],com[maxn];

void prim(int n) {
	mem0(visit);
	mem0(dist);
	int i,j,to;
	s=t=-1;
	for (i=1;i<=n;i++) {
		int m=-1;
		for (j=1;j<=n;j++) {
			if (!com[j]&&!visit[j]&&dist[j]>m) {
				m=dist[j];
				to=j;
			}
		}
		if (t==to) return;
		s=t;t=to;
		visit[to]=1;
		mcut=m;
		for (j=1;j<=n;j++) {
			if (!com[j]&&!visit[j]) {
				dist[j]+=mp[to][j];
			}
		}
	}
}

int SW(int n) {
	int i,j,cas=n-1,ans=inf;
	mem0(com);
	while (cas--) {
		prim(n);
		ans=min(ans,mcut);
		if (ans==0) return 0;
		com[t]=1;
		for (j=1;j<=n;j++) {
			if (!com[j]) {
				mp[s][j]+=mp[t][j];
				mp[j][s]+=mp[j][t];
			}
		}
	}
	return ans;
}

int main() {
	int n,m,i,x,y,d;
	while (scanf("%d%d",&n,&m)!=EOF) {
		num=0;
		mem0(mp);
		for (i=1;i<=m;i++) {
			scanf("%d%d%d",&x,&y,&d);
			mp[x+1][y+1]+=d;
			mp[y+1][x+1]+=d;
		}
		int ans=SW(n);
		printf("%d\n",ans);
	}
	return 0;
}