HDU 4932 贪心
Miaomiao's Geometry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 191 Accepted Submission(s): 38
Problem Description
There are N point on X-axis . Miaomiao would like to cover them ALL by using segments with same length.
There are 2 limits:
1.A point is convered if there is a segments T , the point is the left end or the right end of T.
2.The length of the intersection of any two segments equals zero.
For example , point 2 is convered by [2 , 4] and not convered by [1 , 3]. [1 , 2] and [2 , 3] are legal segments , [1 , 2] and [3 , 4] are legal segments , but [1 , 3] and [2 , 4] are not (the length of intersection doesn't equals zero), [1 , 3] and [3 , 4] are not(not the same length).
Miaomiao wants to maximum the length of segements , please tell her the maximum length of segments.
For your information , the point can't coincidently at the same position.
There are 2 limits:
1.A point is convered if there is a segments T , the point is the left end or the right end of T.
2.The length of the intersection of any two segments equals zero.
For example , point 2 is convered by [2 , 4] and not convered by [1 , 3]. [1 , 2] and [2 , 3] are legal segments , [1 , 2] and [3 , 4] are legal segments , but [1 , 3] and [2 , 4] are not (the length of intersection doesn't equals zero), [1 , 3] and [3 , 4] are not(not the same length).
Miaomiao wants to maximum the length of segements , please tell her the maximum length of segments.
For your information , the point can't coincidently at the same position.
Input
There are several test cases.
There is a number T ( T <= 50 ) on the first line which shows the number of test cases.
For each test cases , there is a number N ( 3 <= N <= 50 ) on the first line.
On the second line , there are N integers Ai (-1e9 <= Ai <= 1e9) shows the position of each point.
There is a number T ( T <= 50 ) on the first line which shows the number of test cases.
For each test cases , there is a number N ( 3 <= N <= 50 ) on the first line.
On the second line , there are N integers Ai (-1e9 <= Ai <= 1e9) shows the position of each point.
Output
For each test cases , output a real number shows the answser. Please output three digit after the decimal point.
Sample Input
3 3 1 2 3 3 1 2 4 4 1 9 100 10
Sample Output
1.000 2.000 8.000HintFor the first sample , a legal answer is [1,2] [2,3] so the length is 1. For the second sample , a legal answer is [-1,1] [2,4] so the answer is 2. For the thired sample , a legal answer is [-7,1] , [1,9] , [10,18] , [100,108] so the answer is 8.
Source
Recommend
简直奇妙,比赛的时候900多就21个过的...,自己当时没考虑到一条线段能覆盖两个点的情况,说究竟还是自己太弱了,不够细心,还有就是自己太心急了,刚敲完就交了,导致罚时比較多,今后得慢慢改,注意到答案仅仅能是距离或者距离的一半,依次枚举即可,对每一个点仅仅有两种选择,一种是选点左边的线段,一种是选右边的线段,当能选左边的时候一定要选左边的,否则选右边的,如果左右两边都不能选,那么这个线段肯定长了,如果当前枚举的距离为x,那么选左边的条件是A[j]-A[j-1]-vis[j]>=x||A[j]==A[j-1]+x,右边这样的就是一条线段覆盖两个点的情况,vis[j]是上一个点对如今这个区间的影响.
代码例如以下:
#include <iostream>
#include <map>
#include<algorithm>
#include <stack>
#include <string.h>
#include <queue>
#include<cstdio>
using namespace std;
#define INF 5e9+7
typedef long long LL;
int main()
{
//freopen("in.txt","r",stdin);
long long T,N;
double A[100];
double vis[100];
cin>>T;
while(T--)
{
cin>>N;
for(int i=1; i<=N; i++)cin>>A[i];
sort(A+1,A+N+1);
double ans=0;
A[N+1]=INF;
for(int i=1; i<=N-1; i++)
{
memset(vis,0,sizeof(vis));
double x=A[i+1]-A[i];
bool ok1=true,ok2=true;
for(int j=2; j<=N-1; j++)
{
if((A[j]-A[j-1]-vis[j]>=x)||(A[j]==A[j-1]+x))
{
continue;
}
if(A[j+1]-A[j]>=x)
{
vis[j+1]=x;
continue;
}
ok1=false;
break;
}
if(ok1)
{
ans=max(x,ans);
}
memset(vis,0,sizeof(vis));
for(int j=2; j<=N-1; j++)
{
if(A[j]-A[j-1]-vis[j]>=x/2||A[j]==A[j-1]+x/2)
{
continue;
}
if(A[j+1]-A[j]>=x/2)
{
vis[j+1]=x/2;
continue;
}
ok2=false;
break;
}
if(ok2)ans=max(ans,x/2);
}
printf("%.3f\n",ans);
}
return 0;
}
