2016夏季练习——类次小生成树

来源：HDU4081

代码：

次优生成树

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
const int INF = 0x3f3f3f3f;
const double MAXDOUBLE = INF*1.0;
int n;
const int MAXN = 1010;
struct Point{
    double x,y;
}p[MAXN];
int ren[MAXN];
double mp[MAXN][MAXN];
double dist[MAXN];
int pre[MAXN];
int vis[MAXN];
double path[MAXN][MAXN];
bool used[MAXN][MAXN];
vector<int>vec[MAXN];
double dis(Point a,Point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double mst;
void prim(){
    mst=0;
    double dm;
    int dv;
    memset(vis,0,sizeof(vis));
    memset(dist,0,sizeof(dist));
    memset(used,0,sizeof(used));
    memset(path,0,sizeof(path));
    for(int i=1;i<=n;i++){
        dist[i] = mp[1][i];
        pre[i] = 1;
    }
    vis[1] = 1;
    dv=-1;
    for(int i=1;i<n;i++){
        dm = MAXDOUBLE;
        dv = -1;
        for(int j=1;j<=n;j++){
            if(!vis[j]&&dm>dist[j]){
                dm = dist[j];
                dv = j;
            }
        }
        used[pre[dv]][dv] = used[pre[dv]][dv] = true;
        vis[dv]=1;
        mst+=mp[pre[dv]][dv];
        for(int j=1;j<=n;j++){
            if(vis[j]&&j!=dv){
                path[dv][j] = path[j][dv] = max(path[j][pre[dv]],dist[dv]);
            }
            if(!vis[j]&&dist[j]>mp[dv][j]){
                dist[j] = mp[dv][j];
                pre[j] = dv;
            }
        }
    }
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%lf%lf%d",&p[i].x,&p[i].y,&ren[i]);
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                mp[i][j]=dis(p[i],p[j]);
            }
            mp[i][i] = 0.0;
        }
        for(int i=0;i<=n;i++)
            vec[i].clear();
        prim();
        double ans=-MAXDOUBLE;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(i!=j){
                    if(used[i][j]){
                        ans = max(ans,(ren[i]+ren[j])/(mst-mp[i][j]));
                    }
                    else
                    {
                        ans = max(ans,(ren[i]+ren[j])/(mst-path[i][j]));
                    }
                }
            }
        }
        printf("%.2f\n",ans);
    }
	return 0;
}