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

题意：有一条小河，河中有青蛙a和青蛙b。青蛙a呆在石头1上，青蛙b呆在石头2上，已知河中总共有n块石头，编号为1~n，已知所有石头的坐标。现在青蛙a想去找青蛙b玩，求青蛙a所需的最小跳跃距离。

解析：这个题意很简单，我们需要先预处理一下数据，建成一个无向图，然后就是dijkstra算法了，但是注意，这里求的是青蛙a所在的石头到青蛙b所在石头的路径上的最大边的值，所以松弛操作要稍微变一下形。

AC代码：

#include 
#include 
#include 
#include 
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[2]);
    }
    return 0;
}

小编福利：要是想进一步了解sqrt（）的RE原因的话，请点击：Compile Error 之 sqrt() 调用 ^_^