[題目鏈接]

http://poj.org/problem?id=2785

[算法]

中途相遇法

[代碼]

#include <algorithm>  
#include <bitset>  
#include <cctype>  
#include <cerrno>  
#include <clocale>  
#include <cmath>  
#include <complex>  
#include <cstdio>  
#include 
 
<cstdlib>  
#include <cstring>  
#include <ctime>  
#include <deque>  
#include <exception>  
#include <fstream>  
#include <functional>  
#include <limits>  
#include <list>  
#include <map>  
#include <iomanip>  
#include <ios>  
#include 
 
<iosfwd>  
#include <iostream>  
#include <istream>  
#include <ostream>  
#include <queue>  
#include <set>  
#include <sstream>  
#include <stdexcept>  
#include <streambuf>  
#include <string>  
#include <utility>  
#include <vector>  
#include 
 
<cwchar>  
#include <cwctype>  
#include <stack>  
#include <limits.h>
using namespace std;
#define MAXN 4010

int i,j,n,ans,l,r;
vector< int > x,y;
int a[MAXN],b[MAXN],c[MAXN],d[MAXN];

int main() 
{
        
        while (scanf("%d",&n) != EOF)
        {
                for (i = 1; i <= n; i++) scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
                x.clear();
                y.clear();
                for (i = 1; i <= n; i++)
                {
                        for (j = 1; j <= n; j++)
                        {
                                x.push_back(a[i] + b[j]);        
                        }    
                } 
                for (i = 1; i <= n; i++)
                {
                        for (j = 1; j <= n; j++)
                        {
                                y.push_back(c[i] + d[j]);
                        }
                }
                sort(x.begin(),x.end());
                sort(y.begin(),y.end());
                ans = 0;
                for (i = 0; i < x.size(); i++)
                {
                        l = lower_bound(y.begin(),y.end(),-x[i]) - y.begin();
                        r = upper_bound(y.begin(),y.end(),-x[i]) - y.begin();    
                        r--;
                        if (y[r] == -x[i]) ans += r - l + 1;
                        else continue;
                }
                printf("%d\n",ans);
        }
        
        return 0;
    
}

[POJ 2785] 4 Values whose Sum is 0