文章标题HDU 2141：Can you find it?（二分）

最新推荐文章于 2020-04-07 19:31:46 发布

Wang_SF2015

最新推荐文章于 2020-04-07 19:31:46 发布

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本文介绍了一种解决三数组求和问题的有效算法。通过预处理两数组的组合并使用二分查找，将复杂度降至O(N^2)，解决了传统暴力匹配方法效率低下的问题。

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Can you find it?

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

题意：有3个数表，从每个表拿出一个数相加能不能得出X。
分析：暴力枚举，O(N^3)爆炸
枚举前两个数，二分第三个O(N^2logN)貌似也可能爆炸。
怎么办？
先处理后两组数字的和O(N^2)；排序，去重同时计重复次数O(N^2)；之后枚举第一个数，二分后两个数的
O(NlogN)
故复杂度为O(N^2)可以过数据
代码：

#include "iostream"
#include "cstdio"
#include "vector"
#include "algorithm"
#include "fstream"
using namespace std;
int a[505],b[505],c[505];
int sum[505*505];
bool flag;
int x;
bool bs(int a[],int key,int k){
    int lo=0,hi=k-1;
    int mi;
    while (lo<=hi){
        mi=((hi-lo)>>1)+lo;
        if (a[mi]==key){
            flag=true;
            return true;
        }
        else if (a[mi]<key)lo=mi+1;
        else hi=mi-1; 
    }
    //flag=false;
    return false;
} 
int main()
{
    int A,B,C;
    int total=0;
//  freopen("in.txt","r",stdin);
    while (scanf ("%d%d%d",&A,&B,&C)!=EOF){
        total++;
        for (int i=0;i<A;i++){
            scanf ("%d",&a[i]);
        }
        for (int i=0;i<B;i++){
            scanf ("%d",&b[i]);
        }
        for (int i=0;i<C;i++){
            scanf ("%d",&c[i]);
        }
        int len=0;
        for (int i=0;i<A;i++){
            for (int j=0;j<B;j++){
            sum[len++]=a[i]+b[j];
            }
        }
        sort(sum,sum+len);

        int S;
        scanf ("%d",&S);
        int s[S];
        for (int i=0;i<S;i++){
            scanf ("%d",&s[i]);
        }
        printf ("Case %d:\n",total);
        for (int i=0;i<S;i++){
            flag=false;
            for (int j=0;j<C;j++){
                if (bs(sum,s[i]-c[j],len)){
                    printf ("YES\n");
                    break;  
                }
            }
            if (!flag)printf ("NO\n");
            //else printf ("YES\n");    
        }       
    }
 }