UVa1152 4 Values whose Sum is 0 (中途相遇法)

Posted 2020-08-08

链接：http://vjudge.net/problem/36014

分析：先枚举a和b，把所有a+b记录下来放在一个有序数组中，然后枚举c和d，查一查-c-d有多少种方法写成a+b的形式（二分查找）。

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 const int maxn = 4000 + 5;
 6 
 7 int a[maxn], b[maxn], c[maxn], d[maxn];
 8 int ab[maxn * maxn], cd[maxn * maxn];
 9 
10 int main() {
11     int T;
12     scanf("%d", &T);
13     while (T--) {
14         int n;
15         scanf("%d", &n);
16         for (int i = 0; i < n; i++) scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
17         for (int i = 0; i < n; i++)
18             for (int j = 0; j < n; j++) ab[i * n + j] = a[i] + b[j];
19         for (int i = 0; i < n; i++)
20             for (int j = 0; j < n; j++) cd[i * n + j] = -(c[i] + d[j]);
21         sort(cd, cd + n * n);
22         long long ans = 0;
23         for (int i = 0; i < n * n; i++)
24             ans += upper_bound(cd, cd + n * n, ab[i]) - lower_bound(cd, cd + n * n, ab[i]);
25         printf("%lld\n", ans);
26         if(T) printf("\n");
27     }
28     return 0;
29 }