POJ 2253 Frogger （最短路

Frogger

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

题意

就是在最短路径里面找一个最长的边

题解：

改下板子的就好 辣鸡POJ在这里卡精度 G++wa拉好几发 改C++就过了ZZ

AC代码

#include <cstdio>#include <cmath>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define LL long long#define CLR(a,b) memset(a,(b),sizeof(a))const int INF = 0x3f3f3f3f;const LL INFLL = 0x3f3f3f3f3f3f3f3f;const int N = 1e4+10;double mps[N][N];       //邻接矩阵，表示从->的距离double dis[N];       //dst[i] 从到i的距离bool vis[N];      // 标记节点是否被访问int n;double m;struct node {    double x, y;}p[N];void init(){    for(int i = 1; i <= n; i++) {        for(int j = 1;j <= n; j++) mps[i][j] = INF;    }}double distt(double x1, double x2, double y1, double y2){    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}void Dijkstra(int s,int t){    //m = 0.0;    for(int i = 1;i <= n; i++) {        if(i == s) continue;        dis[i] =  mps[s][i];    }    vis[s] = true;    double minx;    int k;    for(int i = 1; i <= n; i++){        k = 0;        minx = INF;        for(int j = 1; j <= n; j++) {            if(!vis[j] && dis[j]<minx) {                minx = dis[j];                k = j;            }        }        if(minx > m) m = minx;        if(k == t) return;        vis[k] = true;        for(int j = 1;j <= n; j++) {            if(!vis[j] && mps[k][j]<dis[j]) {                dis[j] =  mps[k][j];            }        }    }}int main(){    int zz = 0;    while(~scanf("%d",&n),n) {        init(); CLR(vis,false);        for(int i = 1;i <= n; i++) {            scanf("%lf%lf",&p[i].x,&p[i].y);        }        for(int i = 1; i <= n; i++) {            for(int j = i+1; j <= n; j++) {                mps[i][j] = mps[j][i] = distt(p[i].x,p[j].x,p[i].y,p[j].y);            }        }        m = 0;        Dijkstra(1,2);        printf("Scenario #%d\n",++zz);        printf("Frog Distance = %.3lf\n",m);        printf("\n");   }return 0;}