UVA 10600 - ACM Contest and Blackout（次小生成树）

题目链接 https://cn.vjudge.net/problem/UVA-10600

【题意】
给定n个点m条边的无向带权图，输出最小生成树和次小生成树的权值（n<=100）

【思路】
求出最小生成树，然后预处理最小瓶颈路，枚举不在最小生成树中的边即可

#include<bits/stdc++.h>
using namespace std;

const int inf=2e9;
const int maxn=105;
const int maxm=5005;

struct Edge{
    int from,to,dist;
    Edge(int f,int t,int d):from(f),to(t),dist(d){}
    bool operator<(const Edge& e)const{
        return dist<e.dist;
    }
};

int n,m;
int par[maxn],vis[maxm],used[maxn],f[maxn][maxn];
vector<Edge> edges,g[maxn];

int find(int x){ return par[x]==x?x:par[x]=find(par[x]); }

int kruscal(){
    for(int i=0;i<=n;++i){ g[i].clear();par[i]=i; }
    sort(edges.begin(),edges.end());
    int ans=0,cnt=0;
    for(int i=0;i<m;++i){
        Edge& e=edges[i];
        int x=find(e.from);
        int y=find(e.to);
        if(x!=y){
            par[x]=y;
            ans+=e.dist;
            vis[i]=true;
            g[e.from].push_back(Edge(e.from,e.to,e.dist));
            g[e.to].push_back(Edge(e.to,e.from,e.dist));
            if(++cnt==n-1) break;
        }
    }
    return ans;
}

void dfs(int u){
    used[u]=true;
    for(int i=0;i<g[u].size();++i){
        Edge& e=g[u][i];
        int v=e.to;
        if(!used[v]){
            for(int x=1;x<=n;++x){
                if(used[x])
                    f[x][v]=f[v][x]=max(f[u][x],e.dist);
            }
            dfs(v);
        }
    }
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        edges.clear();
        for(int i=0;i<m;++i){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            edges.push_back(Edge(u,v,w));
        }
        memset(vis,0,sizeof(vis));
        int mst=kruscal();
        memset(used,0,sizeof(used));
        memset(f,0,sizeof(f));
        dfs(1);
        int ans=inf;
        for(int i=0;i<m;++i){
            if(!vis[i]){
                Edge& e=edges[i];
                ans=min(ans,mst+e.dist-f[e.from][e.to]);
            }
        }
        printf("%d %d\n",mst,ans);
    }
    return 0;
}