[POJ 2253] Frogger【Dijkstra】

  • 2018-01-03
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Problem:

Time Limit: 1000MS Memory Limit: 65536K

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

Solution:

本题是最短路的变形问题,将路径总和最短改为每段最大值最短,仍然可以用 Dijkstra 贪心求解。

【贪心证明】

令石头 i, j 之间的欧几里得距离为 dis[i][j],石头 i 当前的 Frog Distance 值为 frogDis[i],其中第 i 块石头的值最小。即

  • frogDis[i] < frogDis[j], 其中 j ≠ i

假设 frogDis[i] 不是第 i 块石头的最小 Frog Distance,则有

  • max(frogDis[j], dis[j][i]) < frogDis[i] < frogDis[j], 其中 j ≠ i

这显然与 max 的性质相矛盾,故 frogDis[i] 不会被更新,贪心取之即可。

【华丽的 Q.E.D.(证毕)】

注意事项:

在使用 memset 清空 double 数组时,需要赋正无穷时用 0x43 或更大,需要赋负无穷时用 0xc3 或更大。

Code: O(Tn2)

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#define sqr(x) ((x) * (x))
using namespace std;

int T = 0, n, stx[202], sty[202];
double dis[202][202];
double frogDis[202];
bool vis[202];

int main(){
	while(scanf("%d", &n) && n >= 2){
		for(register int i = 1; i <= n; i++) scanf("%d%d", stx + i, sty + i);
		for(register int i = 1; i <= n; i++){
			dis[i][i] = 0;
			for(register int j = i + 1; j <= n; j++)
				dis[i][j] = dis[j][i] = sqrt((double)(sqr(stx[i] - stx[j]) + sqr(sty[i] - sty[j])));
		}
		memset(frogDis, 0x43, sizeof(frogDis)), frogDis[1] = 0;
		memset(vis, 0, sizeof(vis));
		for(register int i = 1; i <= n; i++){
			int minfd = 0;
			for(register int j = 1; j <= n; j++)
				if(!vis[j] && frogDis[j] < frogDis[minfd]) minfd = j;
			vis[minfd] = 1;  // Find the closest stone to expand
			if(minfd == 2) break;  // Freddy reaches Fiona, I'm shining!
			for(register int j = 1; j <= n; j++)
				frogDis[j] = min(frogDis[j], max(frogDis[minfd], dis[minfd][j]));  // Update the frog distance
		}
		printf("Scenario #%d\nFrog Distance = %.3f\n\n", ++T, frogDis[2]);
	}
	return 0;
}

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