poj2002 hash

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建立一张hash表，将所有的顶点加入该表

枚举正方形相邻两个顶点，求出另外两个顶点的坐标，判断这两个顶点是否在hash表中。

#include <iostream>#include <cstdio>using namespace std;#define TABLE_SIZE 2003int p[1005][2];struct Node {    int x, y;    Node *next;} table[TABLE_SIZE];int hash (int x, int y){    return (x * x + y * y) % TABLE_SIZE;}void insert (int x, int y) {    int key = hash(x, y);    Node * newNode = new Node;    newNode->x = x;    newNode->y = y;    newNode->next = table[key].next;    table[key].next = newNode;}bool search (int x, int y) {    Node * cur = table[hash(x, y)].next;    while (cur != NULL) {         if (cur->x == x && cur->y == y) return true;         cur = cur->next;    }    return false;}int main(){    int n, i, j;    while (scanf ("%d", &n) && n) {        for (i = 0; i < TABLE_SIZE; i++) table[i].next = NULL;        for (i = 0; i < n; i++) {            scanf ("%d %d", &p[i][0], &p[i][1]);            insert (p[i][0], p[i][1]);        }        int x1, y1, x2, y2, ans = 0;        for (i = 0; i < n; i++) {            for (j = i + 1; j < n; j++) {                if (i == j) continue;                x1 = p[i][0] + p[i][1] - p[j][1];                y1 = p[i][1] - p[i][0] + p[j][0];                x2 = p[j][0] + p[i][1] - p[j][1];                y2 = p[j][1] - p[i][0] + p[j][0];                if (search(x1, y1) && search(x2, y2)) ans++;                x1 = p[i][0] - p[i][1] + p[j][1];                y1 = p[i][1] + p[i][0] - p[j][0];                x2 = p[j][0] - p[i][1] + p[j][1];                y2 = p[j][1] + p[i][0] - p[j][0];                if (search(x1, y1) && search(x2, y2)) ans++;            }        }        printf ("%d\n", ans / 4);        for (i = 0; i < TABLE_SIZE; i++) {            while (table[i].next != NULL) {                Node * tmp = table[i].next;                table[i].next = tmp->next;                delete tmp;            }        }    }    return 0;}