USACO TRAINING 1.4.2 Packing Rectangles(状态压缩+枚举)

题意：给你四个矩形，给你6种基本摆放，问你如何摆放使得他们能够不重叠放入一个面积最小的矩形里面，求出这个矩形的面积以及它的所有情况

解题思路：枚举所有情况，情况4和情况5可以合并，状态压缩枚举所有的长宽状态以及位置，就可求得

解题代码：

// Packing Rectangles.cpp : 定义控制台应用程序的入口点。
//

//#include "stdafx.h"

/*
ID: dream.y1
PROG: packrec
LANG: C++
*/

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>

using namespace std;

struct node{
  int x, y ; 
}p[5],q[5],a[10000];
int ans = 100000000;
int sum = 0 ;
int ms1,ms2;
int num[6];
void ok() // 对答案进行添加
{
  if(ms1*ms2 < ans)
  {
    ans = ms1*ms2;
    sum = 1;
    memset(a,0,sizeof(a));
    if(ms1 > ms2)
    {
        a[sum].x = ms2;
        a[sum].y = ms1;
    }
    else 
    {
        a[sum].x = ms1;
        a[sum].y = ms2;
    }
  }else if(ms1*ms2 == ans)
  {
    sum ++;
    if(ms1 > ms2)
    {
        a[sum].x = ms2;
        a[sum].y = ms1;
    }
    else 
    {
        a[sum].x = ms1;
        a[sum].y = ms2;
    }
  }
}
int mx(int x, int y )
{
  if(x > y )
      return x; 
  else return y;
}
int hehe1()
{
    ms1 = 0 ; 
    ms2 = 0;
    for(int i =1;i <= 4;i ++)
    {
      ms1 = mx(ms1,q[i].x);
      ms2 += q[i].y;
    }
    return 1;
}
int hehe2()
{
    ms1 = 0 ; 
    ms2 = 0 ;
    int tempc = 0;
    int tempg = 0;
    for(int i = 1;i <= 3; i ++)
    {
      tempc +=q[num[i]].x;
    }
    for(int i = 1;i <= 3;i ++)
    {
      tempg= mx(tempg,q[num[i]].y);
    }
    ms1 = tempg + q[num[4]].x;
    ms2 = mx(tempc,q[num[4]].y);
    return 1;
}
int hehe3()
{
   ms1 = 0 ; 
   ms2 = 0 ;
   ms1 = mx(q[num[2]].x + q[num[3]].x,q[num[1]].y) + q[num[4]].x;
   ms2 = mx(mx(q[num[2]].y ,q[num[3]].y)+q[num[1]].x,q[num[4]].y);
   return 1;
}
int hehe4()
{
    ms1 =0 ; 
    ms2 =0 ;
    if(q[num[1]].x <= q[num[2]].x)
    {
      ms1  = q[num[2]].x + q[num[3]].x + q[num[4]].x;
      ms2 = mx(mx(q[num[3]].y,q[num[4]].y),q[num[2]].y + q[num[1]].y); 
      return 1;
    }else {
       return 0;
    }
}
int hehe5()
{
    ms1 = 0 ; 
    ms2 = 0 ;
    if(q[num[1]].y < q[num[3]].y)
    {
      return 0;
    }else{
        if(q[num[1]].y - q[num[3]].y >= q[num[4]].y)
        {
          ms1 = mx(mx(q[num[2]].y,q[num[1]].x+q[num[4]].x),q[num[1]].x+q[num[3]].x);
          ms2 = q[num[1]].y + q[num[2]].x;
        }else
        {
           ms1 = mx(mx(q[num[2]].y+q[num[4]].x,q[num[1]].x+q[num[3]].x),q[num[1]].x +q[num[4]].x);
           ms2 = mx(q[num[2]].x+q[num[1]].y,q[num[3]].y+q[num[4]].y);
        }
        return 1;
    }
}
void solve(){
    
    for(int i = 0;i <= 15; i ++ )
    {
        int k = i ;
        for(int j = 1;j <= 4;j ++)
        {
          if(k % 2 == 0)
          {
           q[j] = p[j];
          }
          else 
          {
            q[j].x = p[j].y;
            q[j].y = p[j].x;
          }
          k = k/2; 
        }
      for(int i = 1;i <= 4;i ++)
         num[i] = i ;
      do{
      if(hehe1())
          ok();
      if(hehe2())
          ok();
      if(hehe3())
          ok();
      if(hehe4())
          ok();
     if(hehe5())
          ok();
     /* if(hehe6())
          ok();*/
      }while(next_permutation(num+1,num+5));
    }
    
    
}

int cmp(const void * a, const void *b)
{
  return (*(node *)a).x - (*(node*)b).x;
}
int main()
{
   // freopen("packrec.in","r",stdin);
   // freopen("packrec.out","w",stdout);
  FILE *p1, *p2;
  p1 = fopen("packrec.in","r");
  p2 = fopen("packrec.out","w");
 for(int i =1;i <= 4;i ++)
  {
     fscanf(p1,"%d %d",&p[i].x,&p[i].y);
     if(p[i].x > p[i].y)
     {
       int temp ; 
       temp = p[i].x;
       p[i].x = p[i].y ;
       p[i].y = temp;
     }
  }
  solve();
  qsort(a+1,sum,sizeof(node),cmp);
  fprintf(p2,"%d
",ans);
  fprintf(p2,"%d %d
",a[1].x,a[1].y);
  for(int i = 2;i <= sum;i ++)
  {
      if(a[i].x == a[i-1].x || a[i].y == a[i-1].y)
          continue;
      else fprintf(p2,"%d %d
",a[i].x,a[i].y);
      if(i  ==  sum)
          fclose(p2);
  }
  fclose(p1);
  //fclose(p2);
  //printf("%d %d
",a[sum].x,a[sum].y);
  return 0 ;
}