POJ 1971 —— Parallelogram Counting（ZZ）

#include <iostream>
#include <vector>
#include <cstdio>
 
using namespace std;
const int MAXN = 1001;
const int MAXM = 20000;
const int PRIME = 19997;
 
struct Node {
    int x, y;
}node[MAXN];
 
struct Point {
    int midx, midy, cnt;
};
 
vector <Point> hash[MAXM];
int resCnt;
 
void Hash(int sum, Point pt) 
{
    sum %= PRIME;
    vector<Point>::size_type ix, size = hash[sum].size();
    for(ix = 0; ix < size; ix ++) {
        int midx = hash[sum][ix].midx;
        int midy = hash[sum][ix].midy;
        if(midx == pt.midx && midy == pt.midy) {
            resCnt += hash[sum][ix].cnt;
            hash[sum][ix].cnt += pt.cnt;
            break;
        }
    }
    if(ix == size) 
        hash[sum].push_back(pt);
}
 
int main() 
{
    int cas;
    scanf("%d", &cas);
    while(cas --) {
        int i, j, n;
        scanf("%d", &n);
        for(i = 0; i < n; i ++) {
            scanf("%d %d", &node[i].x, &node[i].y);
        }
        for(i = 0; i < PRIME; i ++) 
            hash[i].clear();
        Point pt;
        resCnt = 0;
        for(i = 0; i < n; i ++) {
            for(j = i + 1; j < n; j ++) {
                pt.midx = node[i].x + node[j].x; //不除2，保证整数存储
                pt.midy = node[i].y + node[j].y;
                pt.cnt = 1;
                int sum = pt.midx + pt.midy;
                if(sum < 0) 
                    sum *= -1;
                Hash(sum, pt);
            }
        }
        printf("%d\n", resCnt);
    }
    return(0);
}

题意：有N个点，求这N个点能组成多少个不同的平行四边形。

思路：平行四边形对角形互相平分，所以枚举两个不同的点，求其中点坐标，如果中点坐标相同，说明能组成平行四边形。具体采用哈希查找。

好像曾经见过类似的题目，千万不要枚举 4 个点。

http://www.chenyajun.com/2010/08/03/4846

我的代码

排序过。。。

#include <iostream>
using namespace std;

struct Node
{
    int x;
    int y;
}a[1024];
Node mid[1024*1024];

int cmp(const void* a, const void* b)
{
    Node* c = (Node*)a;
    Node* d = (Node*)b;
    if(c->x != d->x)
        return c->x < d->x ? -1 : 1;
    return c->y < d->y ? -1 : 1;
}

int main()
{
    int cas, n, i, j, count;
    scanf("%d", &cas);
    while(cas--)
    {
        scanf("%d", &n);
        count = 0;
        for(i = 0;  i < n; ++i)
        {
            scanf("%d%d", &a[i].x, &a[i].y);
            for(j = 0; j < i; ++j)
            {
                mid[count].x = a[i].x + a[j].x;
                mid[count].y = a[i].y + a[j].y;
                count++;
            }
        }
        qsort(mid, count, sizeof(mid[0]), cmp);
        int total = 0, sum = 1;
        bool flag = false;
        //printf("%d %d\n", mid[0].x, mid[0].y);
        for(i = 1; i < count; ++i)
        {
            //printf("%d %d...\n", mid[i].x, mid[i].y);
            if(mid[i].x == mid[i-1].x && mid[i].y == mid[i-1].y)
            {
                sum++;
                flag = true;
            }
            else
            {
                if(flag)
                {
                    total += sum*(sum-1)/2;
                    sum = 1;
                }
                flag = false;
            }
        }
        printf("%d\n", total);
    }
    return 0;
}