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

　　　　——每天在线，欢迎留言谈论。

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-;
 typedef long long ll;
 const int MAXN = 4e3+;
 int numbers[][MAXN],sumAB[MAXN*MAXN];
 int main()
 {
     int t;
     cin>>t;
     while(t--)
     {
         memset(numbers,,sizeof(numbers));
         memset(sumAB,INT_INF,sizeof(sumAB));
         int n,p1=,p2=;
         cin>>n;
         for(int i=;i<n;i++)
             for(int j=;j<;j++)
                 cin>>numbers[j][i];
         for(int i=;i<n;i++)
             for(int j=;j<n;j++)
             sumAB[p1++]=numbers[][i]+numbers[][j];
         sort(sumAB,sumAB+p1);
         int answer=;
         for(int i=;i<n;i++)
             for(int j=;j<n;j++)
         {
             answer+=upper_bound(sumAB,sumAB+p1,-numbers[][i]-numbers[][j])-lower_bound(sumAB,sumAB+p1,-numbers[][i]-numbers[][j]);
         }
         cout<<answer<<endl;
         if(t)
             cout<<endl;
     }
     return ;
 }

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