hdu 2141 Can you find it?

最新推荐文章于 2019-05-05 12:09:59 发布

原创最新推荐文章于 2019-05-05 12:09:59 发布 · 650 阅读

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本文探讨了在给定三个整数序列的情况下，如何高效判断是否存在三数之和等于特定目标值的问题。通过合并前两个序列并使用二分查找的方法，显著提升了算法效率。

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Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 7515 Accepted Submission(s): 1950

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

Sample Output

Case 1:
NO
YES
NO

这题n^3明显超时，n^2logn也超时。我们可以合并第一个和第二个数组，然后枚举第三个数组，对合并后的数组进行二分。

AC代码：
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <cmath>
#include <string>
#include <algorithm>
using namespace std;

int a[505],b[505],c[505],d[300000];
bool BinarySearch(int l,int r,int x)
{
    int mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(d[mid]==x) return true;
        if(d[mid]>x) r=mid-1;
        else l=mid+1;
    }
    return false;
}
int main()
{
    int L,N,M,S,x;
    int ca=0;
    while(cin>>L>>N>>M)
    {
        ca++;
        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);
        int k=0;
        for(int i=0; i<L; i++)
            for(int j=0; j<N; j++)
                d[k++]=a[i]+b[j];
        sort(d,d+k);
        scanf("%d",&S);
        printf("Case %d:\n",ca);
        while(S--)
        {
            scanf("%d",&x);
            int flag=0;
            for(int i=0; i<M; i++)
            {
                if(BinarySearch(0,k,x-c[i]))
                {
                    flag=1;
                    break;
                }
                if(flag) break;
            }
            if(flag) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}