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

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>


using namespace std;

/*
Sample Input
4
1 1 1 1
1 1 1 1
1 1 1 1
2 2 -2 -2

Sample Output
5
*/

vector<int> VA(5000);
vector<int> VB(5000);
vector<int> VC(5000);
vector<int> VD(5000);

int n;

map<long long, vector<int> > CDSum;

vector<long long> CDS;

int ret = 0;


void solve()
{
    //将cd的和预先处理好
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            long long sum = VC[i] + VD[j];
        /*    CDSum[sum].push_back(i);
            CDSum[sum].push_back(j);*/
            CDS.push_back(sum);
        }
    }

    sort(CDS.begin(),CDS.end());

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            long long sum = VA[i] + VB[j];
            //map<long long, vector<int> >::iterator it = CDSum.find(-sum);
            /*if (it != CDSum.end()) {
                ret += it->second.size() / 2;
            }*/
            ret += upper_bound(CDS.begin(), CDS.end(), -sum) - lower_bound(CDS.begin(),CDS.end(),-sum);
        }
    }


    cout << ret << endl;
}

int main()
{
    cin >> n;

    for (int i = 0; i < n; i++) {
        cin >> VA[i] >> VB[i] >> VC[i] >> VD[i];
    }

    solve();

    return 0;
}