poj2253青蛙（可到达路径的单次跳跃最短距离）

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 55388		Accepted: 17455

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

传送门：点击打开链接

题目大意：一只青蛙要从点1到点2，给你n个坐标，问你青蛙要达到终点，单次跳跃的最短距离是多少。

思路：一开始想到二分，感觉很麻烦，然后想到djkstra算法里的dis【】，一般我们用这个dis表示从起点点集到某一个点的最短总距离，现在我们可以用dis来表示，从起点点集到某一个点单次跳跃的最短距离，所以有了

for(int j=1;j<=n;j++){
		if(!vis[j])
		dis[j]=min(dis[j],max(g[p][j],dis[p]));
	}

其实就是用三角形，1，p，j三个点，dis【j】要么是本身，要么是另外两条边最大的那一条。

核心思想就是这样，其他的没什么坑点了。然后上完整代码。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<math.h>
#include<cmath>
#include<time.h>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<numeric>
#define ll long long
using namespace std;
const int maxn=210;
const int INF=0x3f3f3f3f;
struct dian {
	double x,y;
} a[maxn];
double g[maxn][maxn];
double dis[maxn];
int vis[maxn],n;
void djks(){
	for(int i=1;i<=n;i++){
		dis[i]=g[1][i];
	}
	memset(vis,0,sizeof(vis));
	vis[1]=1;
	for(int i=1;i<n;i++){
		double minn=INF;
		int p;
		for(int j=1;j<=n;j++){
			if(!vis[j]&&dis[j]<minn){
				p=j;
				minn=dis[j];
			}
		}
		vis[p]=1;
		for(int j=1;j<=n;j++){
			if(!vis[j])
			dis[j]=min(dis[j],max(g[p][j],dis[p]));//核心 用三角形的思路来松弛
		}
	}
}
int main() {
	int cas=1;
	while(scanf("%d",&n),n) {
		memset(g,INF,sizeof(g));
		for(int i=1; i<=n; i++) {
			scanf("%lf%lf",&a[i].x,&a[i].y);
		}
		for(int i=1; i<=n; i++) {
			for(int j=1; j<=n; j++) {
				double x=a[i].x-a[j].x;
				double y=a[i].y-a[j].y;
				g[i][j]=g[j][i]=pow(x*x+y*y,0.5);
			}
		}
		djks();
		printf("Scenario #%d\n",cas++);
		printf("Frog Distance = %.3f\n\n",dis[2]);
	}
}