poj1679The Unique MST(次小生成树)

题意

n个点m条边，判断最小生成树是否唯一

解题思路

求出次小生成树和最小生成树大小是否相同即可

AC代码

#include<vector>
#include<algorithm>
#include<cstdio>
#include<iostream>
#include<set>
#include<cstring>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<int,pii> PII;
const int maxn = 1e6+5;

int B[maxn];
int N,M;
struct Edge 
{
    int from,to,val;
    bool operator<(const Edge& e)const 
    {
        return val < e.val;
    }
};
Edge E[maxn];
int Find(int n)
{
    return B[n]==n?n:B[n]=Find(B[n]);
}

void init()
{
    for(int i=0;i<=N;i++) B[i] = i;
}
int vis[maxn];
int cnt = 0;

int MST()
{
    sort(E,E+M);
    int ans = 0;
    for(int i=0;i<M;i++){
        int t1 = Find(E[i].from);
        int t2 = Find(E[i].to);
        if(t1!=t2){
            if(t1>t2)swap(t1,t2);
            B[t2] = t1;
            vis[cnt++] = i;
            ans += E[i].val;
        }
        if(cnt==N-1){
            break;
        }
    }
    return ans;
}

int SMT()
{
    int res = 1<<30;
    for(int i=0;i<cnt;i++){
        init();
        int t = 0;
        int ans = 0;
        bool ok = 0;
        for(int j=0;j<M;j++){
            if(j!=vis[i]){
                int t1 = Find(E[j].from);
                int t2 = Find(E[j].to);
                if(t1!=t2){
                    t++;
                    ans += E[j].val;
                    if(t1>t2)swap(t1,t2);
                    B[t2] = t1;
                }
                if(t==N-1){
                    ok = 1;
                    break;
                }
            }
        }
        if(ok) res = min(res,ans);
    }
    if(res==1<<30)return -1;
    return res;
}

int main(int argc, char const *argv[])
{
    int T = 0;
    cin >> T;
    while(T--){
        memset(vis,0,sizeof(vis));
        cnt = 0;
        cin >> N >> M;
        init();
        for(int i=0;i<M;i++){
            cin >> E[i].from >> E[i].to >> E[i].val;
        }
        int t1 = MST();
        int t2 = SMT();
        if(t1==t2){
            cout << "Not Unique!" << endl;
        }else{
            cout << t1 << endl;
        }
    }
    
    return 0;
}