POJ 2253 Frogger(最小生成树)
2017-08-23 22:56
483 查看
题目链接:http://poj.org/problem?id=2253点击打开链接
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
最小生成树 当 前两个点链接的时候跳出 不需要继续生成树
#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
#include <stdio.h>
using namespace std;
struct xjy
{
int x;
int y;
};
struct edge
{
double dis;
int be;
int eend;
bool operator < (const edge &r)const
{
return dis<r.dis;
}
};
vector < xjy > s;
vector < edge> ss;
int pre[222];
int findx(int x)
{
int r=x;
while(pre[r]!=r)
{
r=pre[r];
}
int i=x;int j;
while(pre[i]!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
void join ( int x,int y)
{
int p1=findx(x);
int p2=findx(y);
if(p1!=p2)
pre[p2]=p1;
}
int main()
{
int n;int cnt=1;
while(cin >> n,n)
{
for(int i=0;i<222;i++)
pre[i]=i;
s.clear();
ss.clear();
for(int i=0;i<n;i++)
{
xjy mid;
cin >> mid.x >> mid.y;
s.push_back(mid);
}
for(int i=0;i<s.size();i++)
{
for(int j=i+1;j<s.size();j++)
{
edge mid;
mid.be=i;
mid.eend=j;
mid.dis=sqrt((s[i].x-s[j].x)*(s[i].x-s[j].x)+(s[i].y-s[j].y)*(s[i].y-s[j].y));
ss.push_back(mid);
}
}
sort(ss.begin(),ss.end());
double ans;
for(int i=0;i<=ss.size();i++)
{
//cout << ss[i].be << ss[i].eend << endl;
if(findx(ss[i].be)!=findx(ss[i].eend))
{
join(ss[i].be,ss[
b929
i].eend);
if(findx(1)==findx(0))
{
ans=ss[i].dis;
break;
}
}
}
printf("Scenario #%d\n",cnt++);
printf("Frog Distance = ");
printf("%.3f\n",ans);
printf("\n");
}
}
Frogger
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 48449 | Accepted: 15421 |
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
最小生成树 当 前两个点链接的时候跳出 不需要继续生成树
#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
#include <stdio.h>
using namespace std;
struct xjy
{
int x;
int y;
};
struct edge
{
double dis;
int be;
int eend;
bool operator < (const edge &r)const
{
return dis<r.dis;
}
};
vector < xjy > s;
vector < edge> ss;
int pre[222];
int findx(int x)
{
int r=x;
while(pre[r]!=r)
{
r=pre[r];
}
int i=x;int j;
while(pre[i]!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
void join ( int x,int y)
{
int p1=findx(x);
int p2=findx(y);
if(p1!=p2)
pre[p2]=p1;
}
int main()
{
int n;int cnt=1;
while(cin >> n,n)
{
for(int i=0;i<222;i++)
pre[i]=i;
s.clear();
ss.clear();
for(int i=0;i<n;i++)
{
xjy mid;
cin >> mid.x >> mid.y;
s.push_back(mid);
}
for(int i=0;i<s.size();i++)
{
for(int j=i+1;j<s.size();j++)
{
edge mid;
mid.be=i;
mid.eend=j;
mid.dis=sqrt((s[i].x-s[j].x)*(s[i].x-s[j].x)+(s[i].y-s[j].y)*(s[i].y-s[j].y));
ss.push_back(mid);
}
}
sort(ss.begin(),ss.end());
double ans;
for(int i=0;i<=ss.size();i++)
{
//cout << ss[i].be << ss[i].eend << endl;
if(findx(ss[i].be)!=findx(ss[i].eend))
{
join(ss[i].be,ss[
b929
i].eend);
if(findx(1)==findx(0))
{
ans=ss[i].dis;
break;
}
}
}
printf("Scenario #%d\n",cnt++);
printf("Frog Distance = ");
printf("%.3f\n",ans);
printf("\n");
}
}
相关文章推荐
- POJ 2253 Frogger(最小生成树)
- poj 2253 Frogger 最小生成树
- POJ 2253 Frogger(最小生成树)
- poj2253——Frogger(最小生成树,prim算法)
- POJ 2253-Frogger(最小生成树-给定终点)
- POJ 2253 Frogger 最小生成树
- POJ 2253 Frogger(最小生成树)
- POJ 2253 Frogger (最小生成树 or 最短路变形)
- POJ2253--Frogger--最小生成树Kruskal
- Poj 2253 Frogger【最小生成树(巧解)】
- Poj 2253 Frogger(最小生成树的最大边)
- POJ 2253 Frogger (最小生成树.Prim)
- POJ 2253 Frogger 最小生成树
- poj 2253 Frogger【最小生成树变形】【kruskal】
- poj 2253 Frogger(最小生成树)
- PoJ 2253 Prim 最小生成树
- poj 2253 最小生成树 kruskal
- POJ-2253 Frogger dijsktra查找间隔最小的路径
- POJ 2253 Frogger (求每条路径中最大值的最小值,Dijkstra变形)
- POJ 2253:Frogger 求每一条路径最大值里面的最小值