poj 2785 4 Values whose Sum is 0

地址 http://poj.org/problem?id=2785

解法

四层遍历肯定不考虑然后考虑使用map 记录 C和D的各种组合的值

然后遍历 a+b的值在搜索map中有无负的相同的绝对值的数

结果还是TLE了

由于poj的C++比较旧没有unordered_map 所以只有考虑二分查找了

思路还是相同就是查找map 的代码改成二分查找

 1 #include <iostream>
 2 #include <vector>
 3 #include <algorithm>
 4 #include <map>
 5 
 6 
 7 using namespace std;
 8 
 9 /*
10 Sample Input
11 4
12 1 1 1 1
13 1 1 1 1
14 1 1 1 1
15 2 2 -2 -2
16 
17 Sample Output
18 5
19 */
20 
21 vector<int> VA(5000);
22 vector<int> VB(5000);
23 vector<int> VC(5000);
24 vector<int> VD(5000);
25 
26 int n;
27 
28 map<long long, vector<int> > CDSum;
29 
30 vector<long long> CDS;
31 
32 int ret = 0;
33 
34 
35 void solve()
36 {
37     //将cd的和预先处理好
38     for (int i = 0; i < n; i++) {
39         for (int j = 0; j < n; j++) {
40             long long sum = VC[i] + VD[j];
41         /*    CDSum[sum].push_back(i);
42             CDSum[sum].push_back(j);*/
43             CDS.push_back(sum);
44         }
45     }
46 
47     sort(CDS.begin(),CDS.end());
48 
49     for (int i = 0; i < n; i++) {
50         for (int j = 0; j < n; j++) {
51             long long sum = VA[i] + VB[j];
52             //map<long long, vector<int> >::iterator it = CDSum.find(-sum);
53             /*if (it != CDSum.end()) {
54                 ret += it->second.size() / 2;
55             }*/
56             ret += upper_bound(CDS.begin(), CDS.end(), -sum) - lower_bound(CDS.begin(),CDS.end(),-sum);
57         }
58     }
59 
60 
61     cout << ret << endl;
62 }
63 
64 int main()
65 {
66     cin >> n;
67 
68     for (int i = 0; i < n; i++) {
69         cin >> VA[i] >> VB[i] >> VC[i] >> VD[i];
70     }
71 
72     solve();
73 
74     return 0;
75 }