# POJ 2253 Frogger （dijkstra算法 + 预加工）

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POJ 2253 Frogger （dijkstra算法 + 预处理）

Frogger
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 27020 Accepted: 8797

Description

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and instead reach her by jumping.
Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps.
To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence.
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.

You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone.

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output

For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

```2
0 0
3 4

3
17 4
19 4
18 5

0
```

Sample Output

```Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414
```

Source

Ulm Local 1997

AC代码：

```#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define MAXN 200 + 5
#define INF 123456789

struct Node{ int x, y; };        //顶点结构体
int V;                           //顶点数
Node node[MAXN];
double w[MAXN][MAXN], dis[MAXN]; //w：无向图的邻接矩阵，dis：最短路径
int vis[MAXN];                 //访问数组

double dist(Node a, Node b){          //a，b两点之间的距离
return sqrt((double)((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y)));      //注意一定要把sqrt（）函数的参数强转成double类型，否则会CE
}

void dijkstra(){
memset(vis, 0, sizeof(vis));         //预处理
for(int i=1; i<=V; i++) dis[i] = (i==1) ? 0 : INF;
for(int i=1; i<=V; i++){         //dijkstra
int x, m = INF;
for(int y=1; y<=V; y++)
if(!vis[y] && dis[y] <= m){
x = y;
m = dis[x];
}
vis[x] = 1;
for(int y=1; y<=V ;y++) dis[y] = min(dis[y], max(dis[x], w[x][y]));         //变形之后的松弛操作
}
}

int main(){
#ifdef sxk
freopen("in.txt", "r", stdin);
#endif // sxk

int t = 0;
while(scanf("%d", &V)!=EOF && V){
for(int i=1; i<=V; i++)  scanf("%d%d", &node[i].x, &node[i].y);
for(int i=1; i<=V; i++)              //处理数据，建无向图
for(int j=i; j<=V; j++)
w[i][j] = w[j][i] = dist(node[i], node[j]);

dijkstra();          //处理

if(t) printf("\n");     //每个样例之间有空一行，但是最后一个样例后面没有！！！
printf("Scenario #%d\nFrog Distance = %.3lf\n", ++t, dis);
}
return 0;
}
```