POJ-2253 Frogger---最短路变形&&最大边的最小值

题目链接：

https://vjudge.net/problem/POJ-2253

题目大意：

青蛙A想访问青蛙B，必须跳着石头过去，不幸的是，B所在的石头太远了，需要借助其他的石头，求从A到B的路径中，青蛙最少需要的跳跃能力是多远

思路：

理清题意，这里规定的是每条路中的最大边为青蛙需要的跳跃能力，要求这个跳跃能力的最小值，思路和POJ2263一样POJ2263求的是最小边的最大值，这里求的是最大边的最小值，同样是松弛方程改变一下就可以，三种算法均可，这里附上Floyd和dijkstra

Floyd：188ms

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #include<sstream>
 #define MEM(a, b) memset(a, b, sizeof(a));
 using namespace std;
 typedef long long ll;
 const int maxn =  + ;
 const int INF = 0x3f3f3f3f;
 int T, n, m, cases, tot;
 double Map[maxn][maxn];
 struct node
 {
     double x, y;
 };
 node a[maxn];
 int main()
 {
     while(cin >> n && n)
     {
         MEM(a, );
         MEM(Map, );
         for(int i = ; i <= n; i++)
         {
             cin >> a[i].x >> a[i].y;
         }
         for(int i = ; i <= n; i++)
         {
             for(int j = ; j <= n; j++)
                 Map[i][j] = sqrt((a[i].x - a[j].x) * (a[i].x - a[j].x) + (a[i].y - a[j].y) * (a[i].y - a[j].y));
         }
         for(int k = ; k <= n; k++)
         {
             for(int i = ; i <= n; i++)
             {
                 for(int j = ; j <= n; j++)
                 {
                     Map[i][j] = min(Map[i][j], max(Map[i][k], Map[k][j]));
                 }
             }
         }
         printf("Scenario #%d\n", ++cases);
         printf("Frog Distance = %.3f\n\n", Map[][]);
     }
     return ;
 }

dijkstra：47ms

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #include<sstream>
 #define MEM(a, b) memset(a, b, sizeof(a));
 using namespace std;
 typedef long long ll;
 const int maxn =  + ;
 const int INF = 0x3f3f3f3f;
 int T, n, m, cases, tot;
 double Map[maxn][maxn];
 struct node
 {
     double x, y;
 };
 node a[maxn];
 bool v[maxn];
 double d[maxn];
 void dijkstra(int u)
 {
     memset(v, , sizeof(v));
     for(int i = ; i <= n; i++)d[i] = INF;
     d[u] = ;
     for(int i = ; i <= n; i++)
     {
         int x, m = INF;
         for(int i = ; i <= n; i++)if(!v[i] && d[i] <= m)m = d[x = i];//找最小值
         v[x] = ;
         for(int i = ; i <= n; i++)d[i] = min(d[i], max(d[x], Map[x][i]));
     }
 }
 int main()
 {
     while(cin >> n && n)
     {
         MEM(a, );
         MEM(Map, );
         for(int i = ; i <= n; i++)
         {
             cin >> a[i].x >> a[i].y;
         }
         for(int i = ; i <= n; i++)
         {
             for(int j = ; j <= n; j++)
                 Map[i][j] = sqrt((a[i].x - a[j].x) * (a[i].x - a[j].x) + (a[i].y - a[j].y) * (a[i].y - a[j].y));
         }
         dijkstra();
         printf("Scenario #%d\n", ++cases);
         printf("Frog Distance = %.3f\n\n", d[]);
     }
     return ;
 }