POJ 2253

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参考：http://www.cnblogs.com/tanhehe/p/3169865.html

另外请参考：http://blog.csdn.net/PKU_ZZY/article/details/52434239

#include<cstdio>#include<cmath>#define N 205#define INF 1e60double max(double a,double b){return a>b?a:b;}double edge[N][N],d[N];struct Point{int x,y;}point[N];int n;bool vis[N];double dist(Point a,Point b){return sqrt( (b.y-a.y)*(b.y-a.y) + (b.x-a.x)*(b.x-a.x) );} void init(){for(int i=1;i<=n;i++){if(i==1) d[1]=0.0 , vis[1]=1;else d[i]=dist(point[1],point[i]) , vis[i]=0;for(int j=i;j<=n;j++){edge[i][j]=edge[j][i]=dist(point[i],point[j]);}}}double dijkstra(){for(int i=2;i<=n;i++){double min=INF;int x;for(int j=1;j<=n;j++) if(!vis[j] && d[j] <= min) min=d[(x=j)];//找集合Q( Q=G.V - S )里d[x]最小的那个点x vis[x]=1;//把点x放进集合S里 for(int y=1;y<=n;y++)//把在集合Q里所有与点x相邻的点都找出来松弛,因为这里青蛙可以在任意来两石头间跳，所以直接遍历 G.V - S 即可 {if(!vis[y]){double tmp = max(d[x], edge[x][y]);if(d[y] > tmp) d[y] = tmp;}}}return d[2];}int main(){int kase=0;while(scanf("%d",&n) && n!=0){for(int i=1;i<=n;i++) scanf("%d%d",&point[i].x,&point[i].y);init();printf("Scenario #%d\n",++kase);printf("Frog Distance = %.3f\n\n",dijkstra());}}

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