HDOJ 4739 - Zhuge Liang's Mines 暴力DFS

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题意:

给了N个点(N<=20)..问最多能组成多少个正方形(要求边平行与坐标轴)

题解:

直接暴力DFS就行了..每次确定一个正方形.再深入..更新答案..

Program:

#include<iostream>#include<stack>#include<queue>#include<stdio.h>#include<algorithm>#include<string.h>#include<cmath>#define ll long long#define oo 1000000007#define eps 1e-5#define MAXN 3005#define MAXM 3000005using namespace std;   struct node{      int x,y;}P[32]; bool used[32];int num,ans,n;void dfs(int i){      int j,k,h,w1,w2;      ans=max(ans,num);      for (;i<n;i++)      {            if (used[i]) continue;            used[i]=true;              for (j=i+1;j<n;j++)            {                   if (used[j] || P[i].y==P[j].y || P[i].x!=P[j].x) continue;                   used[j]=true;                   h=abs(P[i].y-P[j].y),w1=-1;                   for (w1=0;w1<n;w1++)                       if (P[w1].x==P[i].x+h && P[w1].y==P[i].y && !used[w1]) break;                   if (w1==n) { used[j]=false; continue; }                      used[w1]=true;                   for (w2=0;w2<n;w2++)                       if (P[w2].x==P[j].x+h && P[w2].y==P[j].y && !used[w2]) break;                   if (w2==n){ used[j]=used[w1]=false; continue; }                        used[w2]=true;                   num+=4;                                          dfs(i+1);                          used[j]=used[w1]=used[w2]=false;                     num-=4;               }            used[i]=false;      }      return;}int main(){       int i;         while (~scanf("%d",&n) && n!=-1)      {                 for (i=0;i<n;i++) scanf("%d%d",&P[i].x,&P[i].y);                ans=num=0,memset(used,false,sizeof(used));               dfs(0);               printf("%d\n",ans);      }      return 0;}