http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2224

题意:n个木桩,给出他们的坐标,一些木桩之间连着m个篱笆,组成了多个封闭区域,每个封闭区域里面有猫,要将猫救出来,至少拆多长的篱笆

也就是n个顶点、m条边的图,要使图中没有圈,需要去掉的边的权值至少为多少?

题解:

最小生成树+并查集

求图的最大生成树,总权值-生成树的总权值就是答案。

既然去掉的最小权值不好求,那就求留下的最大权值。

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,l,r,u,v;
int cnt,flag,temp;
double ans,sum;
int pre[N],a[N];
char str;
struct point{
    int x,y;
}p[N];
struct node{
    int u,v;
    double w;
    bool operator<(const node&S)const{
        return w>S.w;
    }
}e[N];
double dist(int i,int j){
    return sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y));
}
int find(int x){return pre[x]==x?x:pre[x]=find(pre[x]);}
bool marge(int u,int v){
    int tu=find(u);
    int tv=find(v);
    if(tu!=tv){
        pre[tu]=tv;
        return 1;
    }
    return 0;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        pre[i]=i;
    }
    for(int i=1;i<=n;i++){
        scanf("%d%d",&p[i].x,&p[i].y);
    }
    for(int i=1;i<=m;i++){
        scanf("%d%d",&u,&v);
        e[++cnt].u=u;
        e[cnt].v=v;
        e[cnt].w=dist(u,v);
        sum+=e[cnt].w;
    }
    sort(e+1,e+cnt+1);
    for(int i=1;i<=cnt;i++){
        if(marge(e[i].u,e[i].v)){
            ans+=e[i].w;
        }
    }
    //cout<<sum-ans<<endl;
    printf("%.3f\n",sum-ans);
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}