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  • HDU4081 Qin Shi Huang's National Road System

    先求最小生成树

    再遍历每一对顶点,如果该顶点之间的边属于最小生成树,则剪掉这对顶点在最小生成树里的最长路径

    否则直接剪掉连接这对顶点的边~

    用prim算法求最小生成树最长路径的模板~

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    #include<cmath>
    using namespace std;
    const int maxn=1014;
    const int inf=1e9;
    double g[maxn][maxn],d[maxn],w[maxn];
    int visit[maxn],N,S,P,pre[maxn];
    double len[maxn][maxn];//表示点i,j在最小生成树上的最长路径 
    double c[maxn][maxn];
    void init () {
        for (int i=0;i<maxn;i++)
        for (int j=0;j<maxn;j++)
        g[i][j]=inf,c[i][j]=0,len[i][j]=0;
    }
    double prim (int s) {
        fill (d,d+maxn,inf);
        fill (visit,visit+maxn,0);
        for (int i=1;i<=N;i++) pre[i]=i;
        d[s]=0;
        double ans=0;
        for (int i=1;i<=N;i++) {
            int u=-1,min=inf;
            for (int j=1;j<=N;j++)
            if (!visit[j]&&d[j]<min) {
                u=j;
                min=d[j];
            }
            if (u==-1) return -1;
            visit[u]=1;
            ans+=d[u];
            if (pre[u]!=u) c[u][pre[u]]=c[pre[u]][u]=1;
            for (int v=1;v<=N;v++)
            if (!visit[v]&&g[u][v]!=inf&&g[u][v]<d[v]) {
                d[v]=g[u][v];
                pre[v]=u;
            } 
            for (int v=1;v<=N;v++)
            if (visit[v]&&u!=v) len[u][v]=len[v][u]=max(len[pre[u]][v],d[u]); 
        }
        return ans;
    }
    struct node {
        double x,y;
    }Node[maxn];
    int main () {
        int T;
        scanf ("%d",&T);
        while (T--) {
            scanf("%d",&N);
            for (int i=1;i<=N;i++) scanf ("%lf %lf %lf",&Node[i].x,&Node[i].y,&w[i]);
            init ();
            for (int i=1;i<=N;i++)
            for (int j=i+1;j<=N;j++)
            g[i][j]=g[j][i]=sqrt((Node[i].x-Node[j].x)*(Node[i].x-Node[j].x)+(Node[i].y-Node[j].y)*(Node[i].y-Node[j].y));
            double mst=prim(1);
            double Max=0;
            for (int i=1;i<=N;i++)
            for (int j=i+1;j<=N;j++) 
            if (!c[i][j]) Max=max(Max,(w[i]+w[j])/(mst-len[i][j]));
            else Max=max(Max,(w[i]+w[j])/(mst-g[i][j]));
            printf ("%.2f
    ",Max);
        } 
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zhanglichen/p/12318743.html
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