POJ 2253 Frogger
2015-11-15 17:01
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题目:
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
Sample Output
Source
Ulm Local 1997
思路:
最短路中的最长路,这题看了很久还是没有看懂题意,最后是看别人的题解才懂的
AC的代码:
Frogger
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 31730 | Accepted: 10223 |
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
思路:
最短路中的最长路,这题看了很久还是没有看懂题意,最后是看别人的题解才懂的
AC的代码:
#include <iostream> #include <cstring> #include <stdio.h> #include <cmath> #define INF 0x3f3f3f3f using namespace std; const int maxn=200+5; double mp[maxn][maxn]; int vis[maxn]; double dis[maxn]; int n; struct tmp{ int x,y; }a[maxn]; double getlen(tmp aa,tmp b){ double tmp=sqrt(1.0*(aa.x-b.x)*(aa.x-b.x)+(aa.y-b.y)*(aa.y-b.y)); return tmp; } void dij(){ double maxx=0; int xx; memset(vis,0,sizeof(vis)); memset(dis,INF,sizeof(dis)); for(int i=1;i<=n;i++) dis[i]=mp[1][i]; vis[1]=1; dis[1]=0; for(int i=1;i<n;i++){ double minn=INF; for(int j=1;j<=n;j++){ if(!vis[j] && dis[j]<minn){ minn=dis[j]; xx=j; } } if(minn==INF)break; vis[xx]=1; for(int j=1;j<=n;j++){ double tmp=max(mp[xx][j],dis[xx]); if(!vis[j]&& dis[j]>tmp){ dis[j]=tmp; } } } } int main(){ int cnt=0; while(cin>>n && n){ cnt++; for(int i=1;i<=n;i++){ cin>>a[i].x>>a[i].y; } memset(mp,INF,sizeof(mp)); for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ mp[i][j]=mp[j][i]=getlen(a[i],a[j]); } } dij(); cout<<"Scenario #"<<cnt<<endl; printf("Frog Distance = %.3f\n\n",dis[2]); } return 0; }
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