POJ 2253 Frogger(并查集+二分)
2018-02-26 21:51
459 查看
Frogger
DescriptionFreddy 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.
InputThe 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.OutputFor 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 Input2
0 0
3 4
3
17 4
19 4
18 5
0
Sample OutputScenario #1
Frog Distance = 5.000
Scenario #2
Frog Distance = 1.414题解:
二分其值,将小于假定值的边连接的两点加入并查集。
最后判断1,2是否在同一并查集中
代码:
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int n;
double x[202],y[202];
double rode[202][202];
int par[202],rank1[202];
void init()
{
for(int i=0; i<202; i++)
par[i]=i,rank1[i]=0;
}
int find(int x)
{
if(x==par[x])return x;
return par[x]=find(par[x]);
}
void unite(int x,int y)
{
x=find(x);
y=find(y);
if(x==y)return;
if(rank1[x]<rank1[y])
par[x]=y;
else
{
par[y]=x;
if(rank1[x]==rank1[y])
rank1[x]++;
}
}
bool Ku(double L)
{
init();
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
{
if(rode[i][j]<=L)
unite(i,j);
}
return find(1)==find(2);
}
int main()
{
int cas=0;
while(~scanf("%d",&n))
{
if(!n)break;
int i,j;
for(i=1; i<=n; i++)
scanf("%lf%lf",&x[i],&y[i]);
for(i=1; i<=n; i++)
{
for(j=1; j<=n; j++)
{
rode[i][j]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
}
}
double l=0.00,r=1500.00;
while(r-l>0.000001)
{
double m=(l+r)/2;
if(Ku(m))r=m;
else l=m;
}
//if(cas)printf("\n");
printf("Scenario #%d\n",++cas);
printf("Frog Distance = %.3lf\n",r);
printf("\n");
}
return 0;
}
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions:53044 | Accepted: 16824 |
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.
InputThe 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.OutputFor 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 Input2
0 0
3 4
3
17 4
19 4
18 5
0
Sample OutputScenario #1
Frog Distance = 5.000
Scenario #2
Frog Distance = 1.414题解:
二分其值,将小于假定值的边连接的两点加入并查集。
最后判断1,2是否在同一并查集中
代码:
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int n;
double x[202],y[202];
double rode[202][202];
int par[202],rank1[202];
void init()
{
for(int i=0; i<202; i++)
par[i]=i,rank1[i]=0;
}
int find(int x)
{
if(x==par[x])return x;
return par[x]=find(par[x]);
}
void unite(int x,int y)
{
x=find(x);
y=find(y);
if(x==y)return;
if(rank1[x]<rank1[y])
par[x]=y;
else
{
par[y]=x;
if(rank1[x]==rank1[y])
rank1[x]++;
}
}
bool Ku(double L)
{
init();
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
{
if(rode[i][j]<=L)
unite(i,j);
}
return find(1)==find(2);
}
int main()
{
int cas=0;
while(~scanf("%d",&n))
{
if(!n)break;
int i,j;
for(i=1; i<=n; i++)
scanf("%lf%lf",&x[i],&y[i]);
for(i=1; i<=n; i++)
{
for(j=1; j<=n; j++)
{
rode[i][j]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
}
}
double l=0.00,r=1500.00;
while(r-l>0.000001)
{
double m=(l+r)/2;
if(Ku(m))r=m;
else l=m;
}
//if(cas)printf("\n");
printf("Scenario #%d\n",++cas);
printf("Frog Distance = %.3lf\n",r);
printf("\n");
}
return 0;
}
相关文章推荐
- POJ 2253 Frogger(并查集+二分)
- POJ - 2253 Frogger(并查集,二分)
- 【POJ】2253 Frogger 二分+bfs
- 【POJ】2253 - Frogger(二分)
- POJ 2253 Frogger 二分 + 最短路
- [POJ 2253] Frogger [二分答案+搜索]
- POJ 2253 Frogger (floyd, 二分)
- POJ 2253 Frogger
- poj-2253-Frogger
- poj 2253 Frogger(floyd变形)
- POJ 2253 Frogger
- POJ 2253 Frogger
- POJ培训计划2253_Frogger(最短/floyd)
- POJ-2253 Frogger-寻找最长边-bellman
- poj 2253 Frogger
- poj_2253 Frogger(floyd + %f)
- POJ 2253 Frogger【最短路变形——路径上最小的最大权】
- POJ2253——Frogger(Floyd变形)
- poj 2253 Frogger 求从a到b路上的最大值中的最小值
- poj-2253 Frogger(dijkstra变形)