CF#524-C

题目链接

http://codeforces.com/contest/1080/problem/C

题意

　　给出一个n*m规格的矩形方块，这个矩形方块初始状态是严格的黑白相间的，现在有两次操作，第一次操作是将一块区域的矩形全部变为白色，第二次操作是将一块区域的矩形覆盖成黑色，要求的是经过两次操作之后白色方块和黑色方块的数量，矩形的区域用两个坐标表示，分别是左下角坐标和右上角坐标。

思路

代码

 1 #include<iostream>
 2 using namespace std;
 3 
 4 int max(int a, int b)
 5 {
 6     return a > b ? a : b ;
 7 }
 8 
 9 int min(int a, int b)
10 {
11     return a < b ? a : b ;
12 }
13 
14 long long getArea(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4)     //计算相交面积
15 {
16     if((min(x2, x4) - max(x1, x3) >= 0) && (min(y2, y4) - max(y1, y3) >= 0))
17     {
18         return (min(x2, x4) - max(x1, x3) + 1) * (min(y2, y4) - max(y1, y3) + 1);
19     }
20     else
21         return 0;
22 }
23 
24 struct Area
25 {
26     long long wnum, bnum;
27 };
28 
29 Area getNum(int x1, int y1, int x2, int y2) //求矩形区域内的白、黑数量
30 {
31     long long a = (x2 - x1 + 1);
32     long long b = (y2 - y1 + 1);
33     long long sum = a * b;
34     Area ans;
35     if((x1 % 2 == 0 && y1 % 2 == 0) || (x1 % 2 == 1 && y1 % 2 == 1))
36     {
37         ans.bnum = sum / 2;
38         ans.wnum = sum - ans.bnum;
39 
40     }
41     else
42     {
43         ans.wnum = sum / 2;
44         ans.bnum = sum - ans.wnum;
45     }
46     return ans;
47 }
48 
49 int main()
50 {
51     int t;
52     int n, m;
53     int x1, y1, x2, y2, x3, y3, x4, y4;
54     Area first, second, add, ans;
55     cin >> t;
56     while(t--)
57     {
58         cin >> n >> m;
59         ans = getNum(1, 1, m, n);
60 
61         cin >> x1 >> y1 >> x2 >> y2;
62         first = getNum(x1, y1, x2, y2);
63 
64         cin >> x3 >> y3 >> x4 >> y4;
65         second = getNum(x3, y3, x4, y4);
66 
67         if(getArea(x1, y1, x2, y2, x3, y3, x4, y4)) //存在相交
68         {
69             add = getNum(max(x1, x3), max(y1, y3), min(x2, x4), min(y2, y4));
70         }
71         else
72         {
73             add.wnum = add.bnum = 0;
74         }
75 
76         ans.wnum += first.bnum;
77         ans.bnum -= first.bnum;
78 
79         /*相交部分：白→白→黑
80                    黑→白→黑*/
81         ans.wnum -= (add.wnum + add.bnum);
82         ans.bnum += (add.wnum + add.bnum);
83 
84         ans.wnum -= (second.wnum - add.wnum);
85         ans.bnum += (second.wnum - add.wnum);
86         cout << ans.wnum << " " << ans.bnum << endl;
87     }
88     return 0;
89 }