POJ - 2914 Minimum Cut(全局最小割-Stoer_Wagner)

Posted 2021-06-28 Frozen_Guardian

题目链接：点击查看

题目大意：给出一张无向图，要求将其分为两个集合，使得最小割最小

题目分析：算法学习自：https://blog.csdn.net/dingdi3021/article/details/101960508

全局最小割，用类最大生成树实现的，时间复杂度 $O(n^3)$

代码：

// Problem: Minimum Cut
// Contest: Virtual Judge - POJ
// URL: https://vjudge.net/problem/POJ-2914
// Memory Limit: 65 MB
// Time Limit: 10000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#define lowbit(x) x&-x
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)
{
	T f=1;x=0;
	char ch=getchar();
	while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x*=f;
}
template<typename T>
inline void write(T x)
{
	if(x<0){x=~(x-1);putchar('-');}
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
const int inf=0x3f3f3f3f;
const int N=510;
int maze[N][N];
int v[N],wg[N],vis[N];
int SW(int n) {
    int res = -1;
    for(int i = 0; i < n; ++i) v[i] = i; //点标号的马甲，一开始都是自己。
    while(n > 1){
        memset(vis,0,sizeof(vis)); //标记数组
        memset(wg,0,sizeof(wg)); //权和数组
        int pre = 0; //起点为0号点
        vis[pre] = 1; //标记起点，虽无实际用处，仅为提醒自己
        for(int i = 1; i < n; ++i){
            int p = -1;
            for(int j = 1; j < n; ++j) if(!vis[v[j]]){ //寻找下个拓展点
                wg[v[j]] += maze[v[pre]][v[j]];  //下个点的权和加上边权
                if(p == -1 || wg[v[j]] > wg[v[p]]) p = j; //寻找wg值最大的点
            }
            vis[v[p]] = 1; //标记下个点已经遍历
            if(i == n - 1){ //最后个点，需要合并
                if(res == -1) res = wg[v[p]];
                else res = min(res,wg[v[p]]); //更新res，取最小割
                for(int j = 0; j < n; ++j){
                    maze[v[pre]][v[j]] += maze[v[p]][v[j]];
                    maze[v[j]][v[pre]] += maze[v[j]][v[p]];
                }
                v[p] = v[--n];
            }
            pre = p;
        }
    }
    return res;
}
int main()
{
#ifndef ONLINE_JUDGE
//	freopen("data.in.txt","r",stdin);
//	freopen("data.out.txt","w",stdout);
#endif
//	ios::sync_with_stdio(false);
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF) {
		memset(maze,0,sizeof(maze));
		while(m--) {
			int u,v,w;
			read(u),read(v),read(w);
			maze[u][v]+=w;
			maze[v][u]+=w;
		}
		printf("%d\\n",SW(n));
	}
	return 0;
}