hdu 6665 （2019 Multi-University Training Contest 8 - 1009）Calabash and Landlord

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3228 Accepted Submission(s): 613

Problem Description

Calabash is the servant of a landlord. The landlord owns a piece of land, which can be regarded as an infinite 2D plane.

One day the landlord set up two orthogonal rectangular-shaped fences on his land. He asked Calabash a simple problem: how many nonempty connected components is my land divided into by these two fences, both finite and infinite? Calabash couldn't answer this simple question. Please help him!

Recall that a connected component is a maximal set of points not occupied by the fences, and every two points in the set are reachable without crossing the fence.

Input

The first line of input consists of a single integer

Output

For each test case, print the answer as an integer in one line.

Sample Input

   3 0 0 1 1 2 2 3 4 1 0 3 2 0 1 2 3 0 0 1 1 0 0 1 1 
 

Sample Output

   3 4 2 
 

   题意:两个矩形，问这两个矩形可以将平面分割成几个部分。 
 

   思路:分析，一共可以分割为2,3,4,5,6这五种情况，4的时候情况比较多，其他四种容易 
 

   判断，所以直接判断其他四种情况即可。（比较菜，没想出来好的方法） 
 

   题目给出x1<x2,y1<y2;所以不会有非矩形的样例 
 

   设结果为ans 
 

   当ans = 6，只有这一种情况 
 

当ans = 3时，有相离，内含的情况

当ans = 2时，重合

当ans = 5时，

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当ans = 4时，（或者还有其他的方式，else了..）

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <iostream>
  4 #include <algorithm>
  5 #define int long long
  6 #define rad rand()%mod+1
  7 using namespace std;
  8 typedef long long ll;
  9 const int maxn = 1e6+5;
 10 const int maxm = 1e3+5;
 11 const int mod = 1e9+7;
 12 
 13 signed main()
 14 {
 15     int x1, x2, x3, x4;
 16     int y1, y2, y3, y4;
 17     int t;
 18     scanf("%lld", &t);
 19     while(t--)
 20     {
 21         scanf("%lld %lld %lld %lld", &x1, &y1, &x2, &y2);
 22         scanf("%lld %lld %lld %lld", &x3, &y3, &x4, &y4);
 23         ///为6的情况
 24         if(x1 < x3&&x4 < x2&&y1> y3&&y2 < y4)
 25         {
 26             printf("6\n");
 27             continue;
 28         }
 29         if(x3 < x1&&x2 < x4&&y3 > y1&&y2 > y4)
 30         {
 31             printf("6\n");
 32             continue;
 33         }
 34         ///相离
 35         if(x2 < x3||x4 < x1||y3 > y2||y1 > y4)
 36         {
 37             printf("3\n");
 38             continue;
 39         }
 40         ///重合
 41         if(x1 == x3&&y1 == y3&&x2 == x4&&y2 == y4)
 42         {
 43             printf("2\n");
 44             continue;
 45         }
 46         ///两条边重合
 47         if(x1 == x3 && x2 == x4&&((y1 < y3&&y4 < y2 )||(y1> y3&&y4 > y2)))
 48         {
 49             printf("4\n");
 50             continue;
 51         }
 52         if(y1 == y3&&y2 == y4&&((x1 < x3&&x4 < x2|| (x3 < x1&&x2 < x4)) ))
 53         {
 54             printf("4\n");
 55             continue;
 56         }
 57         ///内含
 58         if(x1 <= x3&&x4 <= x2&&y4<=y2&&y3>=y1)
 59         {
 60             printf("3\n");
 61             continue;
 62         }
 63         if(x3 <= x1&&x2 <= x4&&y1 >= y3&&y2 <= y4)
 64         {
 65             printf("3\n");
 66             continue;
 67         }
 68         ///有一条边重合  (外)
 69         if(x2 == x3 || x1 == x4||y2 == y3 || y1 == y4)
 70         {
 71             printf("3\n");
 72             continue;
 73         }
 74         ///为5 的情况
 75         if(x1 == x3)
 76         {
 77             if((y1 < y3 && y4 < y2) || (y1 > y3 &&y4 > y2))
 78             {
 79                 printf("5\n");
 80                 continue;
 81             }
 82         }
 83         if(y1 == y3)
 84         {
 85             if((x1 < x3 && x4 < x2)||(x3 < x1&&x2 < x4) )
 86             {
 87                 printf("5\n");
 88                 continue;
 89             }
 90         }
 91         if(x2 == x4)
 92         {
 93             if((y1 < y3 && y4 < y2) || (y1 > y3 && y4 > y2) )
 94             {
 95                 printf("5\n");
 96                 continue;
 97             }
 98         }
 99         if(y2 == y4)
100         {
101             if((x1 < x3 && x4 < x2)||(x1 > x3 && x4 > x2))
102             {
103                 printf("5\n");
104                 continue;
105             }
106         }
107         printf("4\n");
108     }
109     return 0;
110 }
111 
112 /***
113 
114 ***/

hdu 6665 （2019 Multi-University Training Contest 8 - 1009）Calabash and Landlord

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