Hdu 4081 Qin Shi Huang's National Road System

Page 1 of the White Paper. Question: Qin Shihuang wants to build a road project to connect all cities. Then, the Taoist Xu Fu can help him build a path without labor. Two variables are given: Let the population at the beginning and end of the road divide A by the sum of labor in the road other than this B Qin Shihuang wants to make this road repair location, making A/B the biggest idea: use the Minimum Spanning Tree to first generate a full-pass graph and then use DFS to calculate the path of the maximum weight on the unique path between each city and then enumerate each city, SUM-maxcost [U] [V] calculated using the Minimum Spanning Tree

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <vector>#define INF 0x3f3f3f3f#define MAX(a,b) a>b?a:busing namespace std;struct point{    int x,y,p;}P[1005];int set[1005];struct line{    int s,e;    double w;    bool operator < (const line &cmp)const    {        return w<cmp.w;    }}L[1005*1005];double maxcost[1005][1005];double map[1005][1005];vector<int>minmap[1005];bool vis[1005];int find(int x){    while(set[x]!=x)    x=set[x];    return x;}double dis(point tx,point ty){    return sqrt((double)(tx.x-ty.x)*(tx.x-ty.x)+(double)(tx.y-ty.y)*(tx.y-ty.y));}void dfs(int st,int pos){    for(int i=0;i<minmap[pos].size();i++)    {        if(!vis[minmap[pos][i]])        {            vis[minmap[pos][i]]=true;            maxcost[st][minmap[pos][i]]=MAX(maxcost[st][pos],map[pos][minmap[pos][i]]);           // printf("maxcost[%d][%d]=%.4lf\n",st,minmap[pos][i],maxcost[st][minmap[pos][i]]);            dfs(st,minmap[pos][i]);        }    }}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n;        memset(maxcost,0,sizeof(maxcost));        scanf("%d",&n);        int top=0;        for(int i=1;i<=n;i++)        {            scanf("%d%d%d",&P[i].x,&P[i].y,&P[i].p);            for(int j=1;j<i;j++)            {                L[top].s=i;                L[top].e=j;                L[top].w=dis(P[i],P[j]);                map[i][j]=map[j][i]=L[top++].w;            }            minmap[i].clear();            set[i]=i;        }        sort(L,L+top);        double sum=0;        for(int i=0;i<top;i++)        {            int xx=find(L[i].s);            int yy=find(L[i].e);            if(xx!=yy)            {                set[xx]=yy;                sum+=L[i].w;                minmap[L[i].s].push_back(L[i].e);                minmap[L[i].e].push_back(L[i].s);            }        }        //printf("sum = %.4lf\n",sum);        for(int i=1;i<=n;i++)        {            memset(vis,false,sizeof(vis));            dfs(i,i);            maxcost[i][i]=0;        }        double ans=0;        for(int i=1;i<n;i++)        {            for(int j=i+1;j<=n;j++)            {                int tmp=P[i].p+P[j].p;                double tt=1.0*tmp/(sum-maxcost[i][j]);                if(tt>ans)ans=tt;            }        }        printf("%.2lf\n",ans);    }    return 0;}