UVa10803 - Thunder Mountain
2014-07-06 13:55
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题意:坐标系内有n个点,距离小于等于10的任意两点之间有边。求任意两点之间最短路中最大的那个。
思路:Floyd算法,水题。注意两点间距离大于10是没有边的。
思路:Floyd算法,水题。注意两点间距离大于10是没有边的。
#include <iostream> #include <stdio.h> #include <cmath> #include <algorithm> #include <iomanip> #include <cstdlib> #include <string> #include <memory.h> #include <vector> #include <queue> #include <stack> #include <ctype.h> #define INF 1000000000 using namespace std; double townx[110]; double towny[110]; double ans[110][110]; double dist(double x1,double y1,double x2,double y2){ return sqrt( (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2) ); } int main(){ int N; cin>>N; for(int t=1;t<=N;t++){ int n; cin>>n; for(int i=1;i<=n;i++){ scanf("%lf%lf",&townx[i],&towny[i]); } //init for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ ans[i][j]=INF; } } for(int k=1;k<=n;k++){ for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ double dis1=dist(townx[i],towny[i],townx[k],towny[k]); if(dis1>10)dis1=ans[i][k]; double dis2=dist(townx[j],towny[j],townx[k],towny[k]); if(dis2>10)dis2=ans[j][k]; if( (dis1+dis2)<ans[i][j]&&(dis1<10.001||ans[i][k]<INF-1)&&(dis2<10.001||ans[j][k]<INF-1)){ ans[i][j]=(dis1+dis2); } } } } double farest=0; for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ farest=max(farest,ans[i][j]); } } printf("Case #%d:\n",t); if(farest>INF-1){ printf("Send Kurdy\n\n"); }else{ printf("%.4lf\n\n",farest); } } return 0; }
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