hdu4739 Zhuge Liang's Mines --状压dp

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=4739

题意：n个点，求出可以组成的最多的正方形的点数，要求每个点只能用一次，且正方形边平行坐标轴。

分析：把所有点组合，判断是否是平行于坐标轴的正方形，是的话，保存，然后遍历。

#define _CRT_SECURE_NO_DEPRECATE

#include<iostream>
#include<vector>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#include<string>
#include<stdio.h>
#define INF 99999999
#define eps 0.0001
using namespace std;

struct Node
{
int x, y;
};

int n;
Node node[25];
int dp[1 << 20];
int sta[5000];

bool cmp(Node a, Node b)
{
if (a.x == b.x)
return a.y < b.y;
return a.x < b.x;
}

/* 判断是否是平行于坐标轴的正方形 */
bool judge(Node a, Node b, Node c, Node d)
{
if (a.x == b.x&&c.x == d.x&&a.x != d.x)//平行坐标轴
{
if (a.y == c.y&&b.y == d.y)//直角
{
if (b.y - a.y == c.x - a.x)//判断四边相等
{
return 1;
}
}
}
return 0;
}

int main()
{
while (scanf("%d", &n) && n != -1)
{
memset(dp, 0, sizeof(dp));

for (int i = 0; i < n; i++)
scanf("%d %d", &node[i].x, &node[i].y);

sort(node, node + n, cmp);

int num = 0;
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
for (int k = j + 1; k < n; k++)
{
for (int l = k + 1; l < n; l++)
{
//cout << i << " " << j << " " << k << " " << l << endl;
int index = (1 << i) | (1 << j) | (1 << k) | (1 << l);
if (judge(node[i], node[j], node[k], node[l]))
sta[num++] = index;
}
}
}
}

//cout << num << endl;
for (int i = 0; i < (1 << n); i++)
for (int j = 0; j < num; j++)
if ((i&sta[j]) == 0)
dp[i | sta[j]] = max(dp[i | sta[j]], dp[i] + 4);

printf("%d\n", dp[(1 << n) - 1]);
}
return 0;
}