uva10462-（带重边次小生成树）

来源：互联网发布：指针数组初始化为null 编辑：程序博客网时间：2024/05/19 23:04

题解:次小生成树，不过这次带重边，那么可以用邻接表存数据，然后判断一下是不是prim算法里面的那条接上去边即可

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;const int mx = 1e3+5;typedef pair<int,int> P;#define x first#define y secondint dis[mx];int vis[mx];int g[mx][mx];int Max[mx][mx];int pre[mx];vector<P>map[mx];int mark[mx][mx];int n,m;int prim(){    memset(dis,inf,sizeof(dis));    memset(vis,0,sizeof(vis));    dis[1] = 0;    int ans = 0;    for(int i = 1; i <= n; i++){        int minn = inf,x = 0;        for(int j = 1; j <= n; j++)            if(!vis[j]&&minn>dis[j])                minn = dis[x = j];        if(x==0) return -1;        vis[x] = 1;        ans += dis[x];        for(int j = 1; j <= n; j++)            if(vis[j]) Max[x][j] = Max[j][x] = x==j?0:max(Max[pre[x]][j],dis[x]);            else if(dis[j]>g[x][j])                dis[j] = g[x][j],pre[j] = x;    }    return ans;}int check(int sum){    int ans = inf;    for(int u = 1; u <= n; u++)        for(auto d: map[u]){            int v = d.x;            int w = d.y;            if((pre[v]!=u&&pre[u]!=v)||g[u][v]!=w||mark[u][v])                ans = min(ans,sum-Max[u][v]+w);            if((pre[v]==u||pre[u]==v)&&g[u][v]==w)                mark[u][v] = mark[v][u] = 1;        }    return ans;}int main(){    int t;    scanf("%d",&t);    for(int casei = 1; casei <= t ; casei++){        memset(g,inf,sizeof(g));        memset(mark,0,sizeof(mark));        scanf("%d%d",&n,&m);        for(int i = 1; i <= n; i++)            map[i].clear();        for(int i = 1; i <= m; i++){            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            g[u][v] = g[v][u] = min(g[u][v],w);            if(u>v)                swap(u,v);            map[u].push_back({v,w});        }        printf("Case #%d : ",casei);        int ans = prim();        if(ans==-1)            puts("No way");        else{            ans = check(ans);            ans==inf?puts("No second way"):printf("%d\n",ans);        }    }    return 0;}

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