poj 2253 Frogger(floyd变形)
来源:互联网 发布:mac版nba2k14突然变卡 编辑:程序博客网 时间:2024/06/06 12:37
Frogger
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 48728 Accepted: 15518
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.
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
Source
Ulm Local 1997
tips:求一条路径上的每次jump的最大值。
然后找到所有从1号石头到2号石头路径上最大值的最小值。
类似floyd。
dis[i][j]代表从第i号石头调到第j号石头的每次jump的最大距离。
动态转移方程:
dis[j][k]=min(dis[j][k],max(dis[j][i],dis[i][k]));
#include<iostream>#include<cstring>#include<cstdio>#include<cmath>using namespace std;const int inf=0x3f3f3f3f;//dis[i][j]代表从第i号石头调到第j号石头的最小步数 int n;double dis[222][222];double getd(int x1,int x2,int y1,int y2){return sqrt((x1*1.0-x2)*(x1*1.0-x2)+(y1*1.0-y2)*(y1*1.0-y2)); } struct node{int x,y;};void init(){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){dis[i][j]=(i==j?0:inf);}}}node a[222];void floyd(){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){for(int k=1;k<=n;k++){dis[j][k]=min(dis[j][k],max(dis[j][i],dis[i][k]));}}}}int main(){int cas;while(cin>>n,n){cas++;init();for(int i=1;i<=n;i++){cin>>a[i].x>>a[i].y;}for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){dis[i][j]=getd(a[i].x,a[j].x,a[i].y,a[j].y);} } floyd(); cout<<"Scenario #"<<cas<<endl; printf("Frog Distance = %.3f\n\n",dis[1][2]);}return 0; }
阅读全文
0 0
- poj 2253 Frogger(floyd变形)
- Poj 2253 Frogger(floyd||SPFA的变形)
- POJ 2253 Frogger(最短路--floyd变形)
- POJ 2253 Frogger (Floyd的变形)
- POJ 2253 Frogger(最短路变形,floyd算法)
- POJ 2253 Frogger Floyd
- 【Floyd】-POJ-2253-Frogger
- Frogger poj 2253 floyd
- poj 2253 Frogger (floyd)
- poj 2253 Frogger (floyd)
- POJ-2253 Frogger (Floyd)
- POJ 2253 Frogger 【Floyd】
- 【POJ 2253 Frogger】+ Floyd
- ACM floyd变形 Frogger
- Frogger(Floyd变形)
- Frogger(poj2253)-floyd变形
- Poj 2253 Frogger【kuskal变形】
- POJ 2253 Frogger(dijkstra变形)
- BFS Nightmare
- java多线程实现
- keras on spark
- 开源漏洞扫描器合集
- URLConnection 请求头和返回值编码
- poj 2253 Frogger(floyd变形)
- 把中文长行,格式化成短行
- ubuntu 设置时间和网络时间保持一致
- 注解之使用注解来实现切点和advice(3)
- hdu1017 A Mathematical Curiosity
- docker run 和新建
- 通过私有构造器强化不可实例化的能力
- 第一篇
- 用C++实现二叉树的三种遍历方式