[折半枚举] poj 2785 4 Values whose Sum is 0

Values whose Sum is 0
Time Limit: 15000MS Memory Limit: 228000K
Total Submissions: 17715 Accepted: 5209
Case Time Limit: 5000MS
Description

The SUM problem can be formulated as follows: given four lists A, B, C, D of integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B x C x D are such that a + b + c + d = 0 . In the following, we assume that all lists have the same size n .
Input

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer values (with absolute value as large as 228 ) that belong respectively to A, B, C and D .
Output

For each input file, your program has to write the number quadruplets whose sum is zero.
Sample Input

6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45
Sample Output

5
Hint

Sample Explanation: Indeed, the sum of the five following quadruplets is zero: (-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30, -75, 77), (-32, -54, 56, 30).

给定各有n个整数的四个数列A,B,C,D，要从每个数列中各取1个数，使四个数的和为0。求出这样的组合的个数。当一个数列中有多个相同的数字时，把它们作为不同的数字看待。

从4个数列中选择的话，总共有n^4种情况，所以全部判断一遍不可行。不过把他们对半分成AB和CD来考虑，就可以快速解决了。从两个数列中选择的话，只有n²种组合，所以可以进行枚举。先从A，B中取出a，b后，为了使总和为0，需要从C、D中取出c+d=-(a+b)，因此先将C、D中取数字的n²种方法全部枚举出来，将这些和排好序，这样就可以运用二分搜索了。复杂度是O(n²logn)。

有时候问题的规模很大，无法枚举所有元素的组合，但能够枚举一般元素的组合。此时，将问题拆成两半以后分别枚举，再合并它们的结果，这一方法往往特别有效。

#include <iostream>#include <cstdio>#include <algorithm>#include <map>#include <cstring>#include <set>using namespace std;const int MAXN = 5000;int n,a[MAXN];int A[MAXN],B[MAXN],C[MAXN],D[MAXN];int CD[MAXN * MAXN];void checkinput(){    for(int i = 0;i<n;i++)        printf("%d  ",A[i]);    printf("\n");    for(int i = 0;i<n;i++)        printf("%d  ",B[i]);    printf("\n");    for(int i = 0;i<n;i++)        printf("%d  ",C[i]);    printf("\n");    for(int i = 0;i<n;i++)        printf("%d  ",D[i]);    printf("\n");}int main(){    scanf("%d",&n);    for(int i = 0;i<n;i++)        scanf("%d%d%d%d",&A[i],&B[i],&C[i],&D[i]);    for(int i = 0;i<n;i++)    {        for(int j = 0; j < n;j++)        {            CD[i*n+j] = C[i] + D[j];        }    }    sort(CD,CD+n*n);    long long res = 0;    for(int i = 0; i < n; i++)    {        for(int j = 0; j < n; j++)        {            int cd = -(A[i]+B[j]); //在CD中刚好为-(A[i]+B[i])的值            res += upper_bound(CD,CD+n*n,cd) - lower_bound(CD,CD+n*n,cd);        }    }    printf("%I64d\n",res);    return 0;}