poj2785 4 Values whose Sum is 0

4 Values whose Sum is 0

链接：http://poj.org/problem?id=2785

Time Limit: 15000MS		Memory Limit: 228000K
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).

题意：ABCD四个集合，每个集合选一个数，使之加起来=0；

题解：二分，得到出AB组合， CD组合的所有请况，把得到的和排序，双指针使之加起来=0；

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
// 0--n-1 1   n -- 2n-1 0
const int M = 4005, inf = 1e8;
int ans, a[M], b[M], c[M], d[M], p[M*M], q[M*M];
bool cmp(int a, int b){return a > b;}

int main(){
    int n, cnt = 0;
    scanf("%d", &n);
    for(int j = 1; j <= n; j++)scanf("%d%d%d%d", &a[j], &b[j], &c[j], &d[j]);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)p[++cnt] = a[i] + b[j];
    cnt = 0;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)q[++cnt] = c[i] + d[j];
    sort(p+1, p+1+cnt); sort(q+1, q+1+cnt, cmp);
    p[cnt+1] = q[cnt+1] = inf;
    int st = 1, ed = 1;
    while(st <= cnt || ed <= cnt){

        while(p[st] + q[ed] < 0)st++;
        while(p[st] + q[ed] > 0)ed++;
        while(p[st] + q[ed] == 0){
            int cc = st, dd = ed;
            while(p[st] == p[cc])st++;
            while(q[ed] == q[dd])ed++;
            ans += (st - cc) * (ed - dd);
        }
    }
    printf("%d
", ans);
}