首页 > itarticle > poj2253 题解

poj2253 题解

admin 11月 13, 2020 0

链接:poj2253

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

Sample Output

Scenario #1

Frog Distance = 5.000

Scenario #2

Frog Distance = 1.414

题解

最短路变形，松弛及松弛条件改为max(dis[k],g[k][i])<dis[i]


#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<queue>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
using namespace std;
typedef pair<double,int> l;
const int N=507;
double g[N][N],dis[N];
int vis[N],x[N],y[N],n,cnt;
priority_queue<l,vector<l>,greater<l> >q;

double (int a,int b){
    return sqrt(pow((double)x[a]-x[b],2)+pow((double)y[a]-y[b],2));
}

double dijkstra(int s,int t){
    while(!q.empty())q.pop();
    memset(vis,0,sizeof(vis));
    rep(i,1,n)dis[i]=0x3f3f3f3f;
    q.push(l(dis[s]=0,s));
    while(!q.empty()){
        int k=q.top().second;
        q.pop();
        vis[k]=1;
        rep(i,1,n)if(max(dis[k],g[k][i])<dis[i]){
            dis[i]=max(dis[k],g[k][i]);
            if(!vis[i])q.push(l(dis[i],i));
        }
    }
    return dis[t];
}

int main(){
    while(~scanf("%d",&n)&&n){
        rep(i,1,n)scanf("%d%d",&x[i],&y[i]);
        rep(i,1,n)rep(j,1,i-1)g[i][j]=g[j][i]=len(i,j);
        printf("Scenario #%dnFrog Distance = %.3lfnn",++cnt,dijkstra(1,2));
    }
    return 0;
}