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HDU 4932 貪心

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HDU 4932 貪心

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.
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.
Output For each test cases , output a real number shows the answser. Please output three digit after the decimal point.
Sample Input

Sample Output

1.000
2.000
8.000
HintFor 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 BestCoder Round #4
Recommend hujie | We have carefully selected several similar problems for you: 4934 4933 4931 4930 4929

簡直神奇,比賽的時候900多就21個過的...,自己當時沒考慮到一條線段能覆蓋兩個點的情況,說到底還是自己太弱了,不夠細心,還有就是自己太心急了,剛敲完就交了,導致罰時比較多,今後得慢慢改,注意到答案只能是距離或者距離的一半,依次枚舉就行,對每個點只有兩種選擇,一種是選點左邊的線段,一種是選右邊的線段,當能選左邊的時候一定要選左邊的,否則選右邊的,如果左右兩邊都不能選,那麼這個線段肯定長了,假設當前枚舉的距離為x,那麼選左邊的條件是A[j]-A[j-1]-vis[j]>=x||A[j]==A[j-1]+x,右邊這種就是一條線段覆蓋兩個點的情況,vis[j]是上一個點對現在這個區間的影響.

代碼如下:

#include 
#include 
#include
#include 
#include 
#include 
#include
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)
                {
                    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;
}