2017百度之星-1002度度熊的王国战略

度度熊的王国战略

赛后更新

没写出来，不更新了。。。

更新：sw算法，蜜汁T，可能姿势不够骚

更新：这是个假题。。。神tm的并查集。。。。题意是求让至少一个点脱离联通块至少需要的代价。。。。。判断联通输出边权和最小的一个点，如果不联通输出0

AC代码：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define ll long long
#define endl ("\n")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ft (frist)
#define sd (second)
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

int pre[N],n,m,wi[N];
void init(int n){
    for(int i=0; i<=n; ++i) pre[i]=i;
}
int finds(int x){
    return pre[x]==x?x:finds(pre[x]);
}
void unions(int x, int y){
    int fx=finds(x), fy=finds(y);
    pre[fy]=fx;
}

int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    while(cin>>n>>m){
        init(n);mem(wi,0);
        for(int i=1; i<=m; ++i){
            int u,v,w; cin>>u>>v>>w;
            if(u==v) continue;
            wi[u]+=w,wi[v]+=w;
            unions(u,v);
        }
        sort(wi+1,wi+1+n);
        int flag=0;
        for(int i=1; i<=n; ++i){
            if(finds(i)==i) flag++;
        }
        if(flag==1) cout<<wi[1]<<endl;
        else cout<<"0\n";
    }
    return 0;
}