poj2253—Frogger （并查集最短路）_并查集求最短路-优快云博客

本文链接：https://blog.youkuaiyun.com/c_circle/article/details/77587568

本文介绍了一个经典的计算问题——Frogger游戏中的最短跳跃路径问题。目标是找到两块石头之间的最小跳跃距离，使得青蛙能从一块石头跳到另一块石头。通过使用并查集和排序算法，可以有效地解决这个问题。

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Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28514		Accepted: 9263

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

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

题意：给n个点，然后n个坐标，前两个分别是始末坐标，剩下的是可经过坐标，想让青蛙从起点跳到终点，求青蛙的弹跳距离不能低于多少

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<math.h>
using namespace std;
typedef long long LL;
#define N 21000
struct node
{
    int x,y;
    double dist;
} q[N];
int a[205],b[205],par[205];
int F(int x)
{
    int i=x;
    while(par[x]!=x)
        x=par[x];
    return x;
}
bool cmp(node a,node b)
{
    return a.dist<b.dist;
}
int main()
{
    int t,i,k,j,num=1;
    while(scanf("%d",&t),t)
    {
        for(i=1; i<=t; i++)
            par[i]=i;
        for(i=1; i<=t; i++)
            scanf("%d%d",&a[i],&b[i]);
        k=0;
        for(i=1; i<t; i++)
            for(j=i+1; j<=t; j++)
            {
                ++k;
                q[k].x=i;
                q[k].y=j;
                q[k].dist=sqrt((a[j]-a[i])*(a[j]-a[i])+(b[j]-b[i])*(b[j]-b[i]))*1.0;
            }
        sort(q+1,q+1+k,cmp);
        double sum;
        for(i=1; i<=k; i++)
        {
            int aa=F(q[i].x);
            int bb=F(q[i].y);
            if(aa!=bb)
            {
                par[bb]=aa;
                sum=q[i].dist;
            }
            if(F(2)==F(1))
                break;
        }
        printf("Scenario #%d\nFrog Distance = %.3f\n\n",num++,sum);
    }
    return 0;
}