hdu 2141 Can you find it? 【时间优化+二分】

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 16411    Accepted Submission(s): 4172

Problem Description

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

 

Input

There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

 

Output

For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".

 

Sample Input

   
   
   
   

    
    
    
    3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1:
NO
YES
NO
   
   
   
   

这个题目数据咋一看还真不是一般变态，1000*500*500*log（500）的算法与1000*500*500*500基本没啥区别，这样肯定TLE。

怎么办？

办法肯定是有的，为什么会超呢？

不就是要对A，B，C三个数组进行1000次询问吗？我要频繁地对这个A，B，C数组进行访问，而A，B，C数组的规模不是很大，我何不另外开一个数组来存A+B可能出现的值呢？然后只要在保存A+B的这个数组里面二分搜索Sum-C[i]的值就OK了，这样我的复杂度就只有1000*500*log（500*500）了，这样就足够保证AC了...

这个时候，我相信你肯定恍然大悟，不要看下面的代码了，还说什么，直接敲吧。

#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 500+5;
int L,M,N,S;
int A[maxn],B[maxn],AB[maxn*maxn],C[maxn],X;
int main () {
    //freopen("input.in","r",stdin);
    int cas = 0;
    while(~scanf("%d%d%d",&L,&M,&N))
    {
        for(int i = 0;i < L;i++) scanf("%d",&A[i]);
        for(int i = 0;i < N;i++) scanf("%d",&B[i]);
        for(int i = 0;i < M;i++) scanf("%d",&C[i]);
        sort(C,C+M);

        for(int i = 0,pos = 0;i < L;i++)
            for(int j = 0;j < N;j++)
                AB[pos++] = A[i]+B[j];
        sort(AB,AB+L*N);
        int LN = unique(AB,AB+L*N) - AB;

        printf("Case %d:\n",++cas);
        scanf("%d",&S);
        for(int xx = 0;xx < S;xx++)
        {
            scanf("%d",&X);
            int sum,c,suc;
            for(int i = 0;i < M;i++)
            {
                sum = X-C[i];
                suc = binary_search(AB,AB+LN,sum);
                if(suc) break;
            }
            printf("%s\n",suc?"YES":"NO");
        }
    }
    return 0;
}