UVA 1152 4 Values whose Sum is 0（中途相遇法）

题意：从四个集合各选一个数，使和等于0，问有多少种选法。

分析：求出来所有ai + bi，在里面找所有等于ci + di的个数。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 40000 + 10;
const int MAXT = 16000000 + 10;
using namespace std;
int a[MAXN], b[MAXN], c[MAXN], d[MAXN];
int sum[MAXT];
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
            int aa, bb, cc, dd;
            scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
        }
        int cnt = 0;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                sum[cnt++] = a[i] + b[j];
            }
        }
        sort(sum, sum + cnt);
        int ans = 0;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                int t = -(c[i] + d[j]);
                ans += upper_bound(sum, sum + cnt, t) - lower_bound(sum, sum + cnt, t);//sum中可能有许多与t相等的数，都要计算上
            }
        }
        printf("%d\n", ans);
        if(T) printf("\n");
    }
    return 0;
}