HDU1875 通畅工程再续 最小生成树

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;struct Island{    int xi,yi,father,z;}island[101];struct Bridge{    int a,b;    double price;}bridge[10010];int cmp(Bridge a,Bridge b){    return a.price < b.price;}int find(int x){    int t;    int child = x;    while(x != island[x].father)    {        x = island[x].father;    }        while(x != child)        {            t=island[child].father;            island[child].father = x;            child = t;        }       return x;}int merge(int x,int y){    int a =find(x);    int b =find(y);    if(a != b)    {        island[a].father = b;        return 1;    }    return 0;   } int main(){    int n,m,i,j,k,g,t; //g 是用来优化最后一步  如果当所有的岛都联通之后 停止 造桥     double ans;    scanf("%d",&t);    while(t--)    {        k = 1;        g =1;        ans = 0.0;        memset(bridge,0,sizeof(bridge));        memset(island,0,sizeof(island));        for(i = 0;i <= 100;i++ )        island[i].father = i;        scanf("%d",&n);        for (i = 1;i <= n;i++)        {            scanf("%d%d",&island[i].xi,&island[i].yi);            island[i].z = i;            for( j = 1;j < i;j++)            {                if(sqrt((island[i].xi - island[j].xi)*(island[i].xi - island[j].xi) +(island[i].yi - island[j].yi)*(island[i].yi - island[j].yi)) <=1000 && sqrt((island[i].xi - island[j].xi)*(island[i].xi - island[j].xi) +(island[i].yi - island[j].yi)*(island[i].yi - island[j].yi) )>= 10)                               {                bridge[k].a = i;                bridge[k].price = sqrt((island[i].xi - island[j].xi)*(island[i].xi - island[j].xi) +(island[i].yi - island[j].yi)*(island[i].yi - island[j].yi) );                bridge[k++].b = j;                 }            }               }               sort(bridge+1,bridge+k,cmp);            for(i = 1;i < k;i++)            {                if (merge(bridge[i].a,bridge[i].b))                {                ans += bridge[i].price;                g++;                }                               if(g == n)                break;            }            g = 0;  //多次利用 而已             for(i = 1;i <= n;i++)            if(island[i].father == i)                g++;                if(g > 1 )            printf("oh!\n");            else            printf("%.1lf\n",ans*100);    }    return 0;}附上网上的方法#include<stdio.h>#include<math.h>#include<string.h>struct Point{       int x,y;} pt[105];double d[105],map[105][105];bool v[105];int c;double Dist(Point a,Point b){       return sqrt( (a.x-b.x)*(a.x-b.x)*1.0+(a.y-b.y)*(a.y-b.y) );}void Graph(){     int i,j;     double t;     memset(map,0,sizeof(map));     for( i=0; i<c-1; i++){          for( j=i+1; j<c; j++){               t=Dist(pt[i],pt[j]);               if( t>=10&&t<=1000){                   map[i][j]=t;                   map[j][i]=t;               }          }     }}double Prim(){    int i,j,pt;    double ret,mim;    memset(d,0,sizeof(d));    memset(v,false,sizeof(v));    pt=0;  v[0]=true;  ret=0;    while( true){           for( i=0; i<c; i++)                 if( !v[i]&&map[pt][i]&&(d[i]>map[pt][i]||d[i]==0) )                    d[i]=map[pt][i];             pt=-1; mim=10000;           for( i=0; i<c; i++){                if( !v[i]&&d[i]&&mim>d[i] ){                    mim=d[i];                    pt=i;                }           }                  v[pt]=true;           if( pt==-1) break;          ret+=mim;     }    for( i=0; i<c; i++)//判断是否联通         if( !v[i])             return -1;     return ret;}int main(){    int t,i;    double ret;    scanf("%d",&t);    while( t--){           scanf("%d",&c);           for( i=0; i<c; i++)                  scanf("%d%d",&pt[i].x,&pt[i].y);            Graph();           ret=Prim();           if( ret==-1)               printf("oh!\n");           else{                ret=ret*100;                printf("%.1lf\n",ret);           }    }    return 0;}