UVa 1152 和为0的4个值（二分查找）

Posted 2020-08-29

 

https://vjudge.net/problem/UVA-1152

题意：给定4个n元素集合A,B,C,D，要求分别从中选取一个元素a,b,c,d，使得a+b+c+d=0。问有多少种取法。

思路：直接暴力枚举的话是会超时的。可以选把a+b的值枚举出来存储，c和d的值也一样并排序，这样就可以在c和d中进行二分查找了。

 1 #include<iostream>  
 2 #include<algorithm>
 3 using namespace std;
 4 
 5 const int maxn = 4000 + 5;
 6 
 7 int n;
 8 int a[maxn], b[maxn], c[maxn], d[maxn];
 9 int s1[25000000], s2[25000000];
10 
11 
12 void solve()
13 {
14     int k = 0;
15     int cnt = 0;
16     for (int i = 0; i < n; i++)
17     {
18         for (int j = 0; j < n; j++)
19         {
20             s1[k] = a[i] + b[j];
21             s2[k] = c[i] + d[j];
22             k++;
23         }
24     }
25     sort(s2, s2 + k);
26     for (int i = 0; i < k; i++)
27     {
28         int t = s1[i];
29         int left = 0, right = k-1;
30         while (left <= right)
31         {
32             int mid = (left + right) / 2;
33             if (s2[mid] == -t)
34             {
35                 cnt++;
36                 int p1 = mid;
37                 int p2 = mid;
38                 //因为有可能存在一样的数，所以前后还需要判断
39                 while (s2[++p1] == -t && p1<k )  cnt++;
40                 while (s2[--p2] == -t && p2>=0)  cnt++;
41                 break;
42             }
43             else if (s2[mid]>-t)  right = mid-1;
44             else left = mid+1;
45         }
46     }
47     cout << cnt << endl;
48 }
49 
50 int main()
51 {
52     //freopen("D:\\txt.txt", "r", stdin);
53     int t;
54     cin >> t;
55     while (t--)
56     {
57         cin >> n;
58         for (int i = 0; i < n; i++)
59         {
60             cin >> a[i] >> b[i] >> c[i] >> d[i];
61         }
62         solve();
63         if (t)  cout << endl;
64     }
65     return 0;
66 }