POJ2785：4 Values whose Sum is 0（二分+暴力）

4 Values whose Sum is 0

Time Limit: 15000MS Memory Limit: 228000KTotal Submissions: 20863 Accepted: 6282Case 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 -4626 -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).

Source

Southwestern Europe 2005

题意：从4列数据每列选1个使之和为0，求可挑选的方案总数。

思路：将第1，2列和第3，4列的和全部求出来，排序，然后二分直至找到和为0。

# include <stdio.h># include <algorithm># define MAXN 4000using namespace std;int a[MAXN][4], b[MAXN*MAXN], c[MAXN*MAXN];int main(){    int n, l, r, sum, tmp, cnt;    while(~scanf("%d",&n))    {        sum = cnt = 0;        for(int i=0; i<n; ++i)            scanf("%d%d%d%d",&a[i][0],&a[i][1],&a[i][2],&a[i][3]);        for(int i=0; i<n; ++i)            for(int j=0; j<n; ++j)            {                b[cnt] = a[i][0] + a[j][1];                c[cnt++] = a[i][2] + a[j][3];            }        sort(b, b+cnt);        sort(c, c+cnt);        for(int i=0; i<cnt; ++i)        {            tmp = 0;            l = 0;            r = cnt-1;            while(l<r)            {                int mid = (l+r)>>1;                if(b[i]+c[mid]<0)                    l = mid + 1;                else                    r = mid;            }            if(r==0&&(b[i]+c[r]!=0) || l==cnt-1&&(b[i]+c[r]!=0))continue;            if(b[i]+c[r]==0) ++tmp;            for(int k=r-1; k>=0&&(b[i]+c[k])==0; --k) ++tmp;//符合情况的c[r]可能扎堆在一起，向前搜索。            for(int k=r+1; k<cnt&&(b[i]+c[k])==0; ++k) ++tmp;//向后搜索。            sum += tmp;            for(;i+1<cnt&&b[i]==b[i+1]; ++i) sum += tmp;//计算相同的b[i]。        }        printf("%d\n",sum);    }    return 0;}