UVA 1152 4 Values whose Sum is 0 (枚举+中途相遇法)(+Java版)(Java手撕快排+二分)

4 Values whose Sum is 0

 题目链接:https://cn.vjudge.net/problem/UVA-1152

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题目大意：

 　　给定4个n(1<=n<=4000)元素的集合 A、B、C、D ，从4个集合中分别选取一个元素a, b,c,d。求满足 a+b+c+d=0的对数。

思路：

　　直接分别枚举 a,b,c,d ，坑定炸了。我们先枚举 a+b并储存，在B、C中枚举找出(-c-d)后进行比较即可。

亮点：

　　由于a+b,中会有值相等的不同组合，如果使用map来存，很遗憾超时(虽然时间限制是9000ms)。

　　在一个有序数组求某元素数出现的个数：n = upper_bound(a,a+n,k)-lower_bound(a,a+n,k) ;

C++ AC代码：

#include <iostream>
#include <cmath>
#include <iostream>
#include <string>
#include <string.h>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
const int INT_INF = 0x3f3f3f3f;
const double EPS = 1e-7;
typedef long long ll;
const int MAXN = 4e3+100;
int numbers[4][MAXN],sumAB[MAXN*MAXN];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        memset(numbers,0,sizeof(numbers));
        memset(sumAB,INT_INF,sizeof(sumAB));
        int n,p1=0,p2=0;
        cin>>n;
        for(int i=0;i<n;i++)
            for(int j=0;j<4;j++)
                cin>>numbers[j][i];
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            sumAB[p1++]=numbers[0][i]+numbers[1][j];
        sort(sumAB,sumAB+p1);
        int answer=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
        {
            answer+=upper_bound(sumAB,sumAB+p1,-numbers[2][i]-numbers[3][j])-lower_bound(sumAB,sumAB+p1,-numbers[2][i]-numbers[3][j]);
        }
        cout<<answer<<endl;
        if(t)
            cout<<endl;
    }
    return 0;
}

Java AC代码：

import java.util.Scanner;
public class Main {
    static Scanner scn = new Scanner(System.in);
    
    static final int MAXN = 4100;
    static int[][] numbers = new int[4][MAXN];
    static int[] sumAB = new int[MAXN*MAXN];
    public static void main(String[] args) {
        int t;   
        t = scn.nextInt();
        while ((t--) > 0) {
            int n;
            n = scn.nextInt();
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < 4; j++) {
                    numbers[j][i] = scn.nextInt();
                }
            }
            int p1 = 0;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    sumAB[p1++] = numbers[0][i] + numbers[1][j];
                }
            }
            Tool.quickSort(sumAB,0,p1-1);
            //输出看看
//            for (int i = 0; i < p1; i++) {
//                System.out.print(sumAB[i] + " ");
//            } //成功
            //开始枚举-c-d
            int answer = 0;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    answer += Tool.uppper(sumAB, 0, p1, -numbers[2][i] - numbers[3][j]) - Tool.lower(sumAB, 0, p1, -numbers[2][i] - numbers[3][j]);
                }
            }
            System.out.println(answer);
            if(t > 0)
                System.out.println();
        }
        System.exit(0);
    }
}
class Tool {
    public static int lower(int[] array, int low, int high, int number) {
        //[low,high)
        int i = low, j = high, m;
        while (i < j) {
            m = i + (j - i) / 2;
            if (array[m] >= number)
                j = m;
            else 
                i = m+1;
        }
        return i;
    }
    public static int uppper(int[] array, int low, int high, int number) {
        int i = low, j = high, m;
        while (i < j) {
            m = i + (j - i) / 2;
            if (array[m] <= number)
                i = m+1;
            else
                j = m;
        }
        return i;
    }
    public static void quickSort(int[] array, int low, int high) {
        if (low < high) {
            int m = partition(array, low, high);
            quickSort(array, low, m-1);
            quickSort(array, m+1, high);
        }
    }
    private static int partition(int[] array, int low, int high) {
        int mNumber = array[low];
        int i = low, j = high;
        while (i < j) {
            while (i < j && array[j] >= mNumber) {--j;}
            array[i] = array[j];
            while (i < j && array[i] <= mNumber) {++i;}
            array[j] = array[i];
        }
        array[i] = mNumber;
        return i;
    }
}

2017-07-30 11:06:43 -> 2017-07-30 13:45:32