POJ 2253 Frogger(青蛙跳石头)

探讨Frogger问题中的最短路径算法实现,通过计算两青蛙间最小最大跳跃距离来寻找最优路径。

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http://poj.org/problem?id=2253

Frogger
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 37886 Accepted: 12196

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

题意:

给定两个青蛙的坐标,A青蛙拜访B青蛙,要求

拜访路程选择的路线上可以跳的最大距离,小于其它任何路线上可以跳的最大距离

如图:


思路:

按照前天做过一类型的题目,找出最短路径下的最大权值边。后来一想,第二组测试数据 ans=2 昂,就卡壳了。

更重要的是你要知道这是最短路径!

其求的就是选择的路线上可以跳的最大距离,小于其它任何路线上可以跳的最大距离


参考CODE:

提交选择C++

#include<stdio.h>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MYDD=1103;
const int INF=0x3f3f3f3f;

struct Q {
	int x,y;
} node[MYDD];
double Dis(Q a,Q b) {/*两点距离*/
	double X=(a.x-b.x)*1.0;
	double Y=(a.y-b.y)*1.0;
	return sqrt(X*X+Y*Y);
}

double dis[MYDD];
bool vis[MYDD];
void dijk(int n) {
	memset(vis,0,sizeof(vis));
	for(int j=0; j<=n; j++)
		dis[j]=INF;
	dis[1]=0;
	for(int j=1; j<=n; j++) {
		int Tp=INF,choose=-1;
		for(int k=1; k<=n; k++) {
			if(!vis[k]&&dis[k]<Tp) {
				Tp=dis[k];
				choose=k;
			}
		}
		vis[choose]=1;
		if(choose==2)	break;/*找到第二只青蛙*/
		for(int k=1; k<=n; k++) {
			if(!vis[k]&&dis[k]>max(dis[choose],Dis(node[choose],node[k])))
				dis[k]=max(dis[choose],Dis(node[k],node[choose]));
		}/*[Error] expected ')' before 'dis' 英语渣渣的 me */
	}
}

int main() {
	int n,kc=1;
	while(scanf("%d",&n)&&n) {
		for(int j=1; j<=n; j++) {
			scanf("%d%d",&node[j].x,&node[j].y);
		}

		dijk(n);
		printf("Scenario #%d\n",kc++);
		printf("Frog Distance = %.3lf\n\n",dis[2]);
	}
	return 0;
}


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