POJ-2253-Frogger

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 42059 Accepted: 13444

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

20 03 4317 419 418 50

Sample Output

Scenario #1Frog Distance = 5.000Scenario #2Frog Distance = 1.414解题思路：这个道题准确的来说不是最短路，也不是最小生成树，定义 f(i,j)为i->j的路径上的最大跳的最小值,那么f(i,j)=min( f(i,j), max(f(i,k),f(k,j)) ),o(n^3)不过用dijkstrea or prime都可以写出来，只要控制好条件。。。然而我的没过心累这是我的代码：看的一个大神的博客才懂的变式dijkstra改变距离定义下dijkstra成立的充分条件

#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>using namespace std;const int MAXN = 205;const int INF = 0x3f3f3f3f;struct Edge{int from;int to;}edge[MAXN];double cost[MAXN][MAXN];int t;void dijkstra(){bool used[MAXN];double dis[MAXN];memset(dis, INF, sizeof(dis));memset(used, false, sizeof(used));for (int i = 0; i < t; i++) {dis[i] = cost[0][i];}dis[0] = 0;used[0] = true;double minDis = INF;while (true) {int v = -1;for (int i = 0; i < t; i++) {if (!used[i] && (v == -1 || dis[v] > dis[i])) {v = i;}}if (v == -1) {break;}used[v] = true;for (int i = 0; i < t; i++) {dis[i] = min(dis[i], max(dis[v], cost[v][i]));}}printf("Frog Distance = %.3f\n\n", dis[1]);}int main(){int caseNum = 1;while(scanf("%d", &t) && t) {int a, b;memset(cost, INF, sizeof(cost));for (int i = 0; i < t; i++) {scanf("%d%d", &edge[i].from, &edge[i].to);}for (int i = 0; i < t; i++) {for (int j = i + 1; j < t; j++) {cost[j][i] = cost[i][j] = sqrt(pow(fabs(edge[i].from - edge[j].from), 2) + pow(fabs(edge[i].to - edge[j].to), 2));}}printf("Scenario #%d\n", caseNum++);dijkstra();}return 0;}
//ps:poj这道题，输出是要用%f.