POJ 2002 Squares(枚举+几何)
2015-03-03 21:03
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思路:先把点排序(为了减少枚举量),然后每次枚举两个点,利用公式计算出另外两个点,在set里面判断存不存在,如果存在就可以构成正方形,这样的计算,一个正方形会被计算两次,所以答案除2即可
代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
int n;
const int N = 1005;
set<pii> vis;
pii p
;
bool judge(int i, int j) {
if (vis.find(make_pair(p[i].second - p[j].second + p[i].first, p[j].first - p[i].first + p[i].second)) == vis.end()) return false;
if (vis.find(make_pair(p[i].second - p[j].second + p[j].first, p[j].first - p[i].first + p[j].second)) == vis.end()) return false;
return true;
}
int main() {
while (~scanf("%d", &n) && n) {
vis.clear();
for (int i = 0; i < n; i++) {
scanf("%d%d", &p[i].first, &p[i].second);
vis.insert(make_pair(p[i].first, p[i].second));
}
sort(p, p + n);
ll ans = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (judge(i, j)) ans++;
}
}
printf("%lld\n", ans / 2);
}
return 0;
}
代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
int n;
const int N = 1005;
set<pii> vis;
pii p
;
bool judge(int i, int j) {
if (vis.find(make_pair(p[i].second - p[j].second + p[i].first, p[j].first - p[i].first + p[i].second)) == vis.end()) return false;
if (vis.find(make_pair(p[i].second - p[j].second + p[j].first, p[j].first - p[i].first + p[j].second)) == vis.end()) return false;
return true;
}
int main() {
while (~scanf("%d", &n) && n) {
vis.clear();
for (int i = 0; i < n; i++) {
scanf("%d%d", &p[i].first, &p[i].second);
vis.insert(make_pair(p[i].first, p[i].second));
}
sort(p, p + n);
ll ans = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (judge(i, j)) ans++;
}
}
printf("%lld\n", ans / 2);
}
return 0;
}
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