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

#include<iostream>
using namespace std;

int max(int a, int b)
{
    return a > b ? a : b ;
}

int min(int a, int b)
{
    return a < b ? a : b ;
}

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
{
    if((min(x2, x4) - max(x1, x3) >= 0) && (min(y2, y4) - max(y1, y3) >= 0))
    {
        return (min(x2, x4) - max(x1, x3) + 1) * (min(y2, y4) - max(y1, y3) + 1);
    }
    else
        return 0;
}

struct Area
{
    long long wnum, bnum;
};

Area getNum(int x1, int y1, int x2, int y2) //求矩形区域内的白、黑数量
{
    long long a = (x2 - x1 + 1);
    long long b = (y2 - y1 + 1);
    long long sum = a * b;
    Area ans;
    if((x1 % 2 == 0 && y1 % 2 == 0) || (x1 % 2 == 1 && y1 % 2 == 1))
    {
        ans.bnum = sum / 2;
        ans.wnum = sum - ans.bnum;

    }
    else
    {
        ans.wnum = sum / 2;
        ans.bnum = sum - ans.wnum;
    }
    return ans;
}

int main()
{
    int t;
    int n, m;
    int x1, y1, x2, y2, x3, y3, x4, y4;
    Area first, second, add, ans;
    cin >> t;
    while(t--)
    {
        cin >> n >> m;
        ans = getNum(1, 1, m, n);

        cin >> x1 >> y1 >> x2 >> y2;
        first = getNum(x1, y1, x2, y2);

        cin >> x3 >> y3 >> x4 >> y4;
        second = getNum(x3, y3, x4, y4);

        if(getArea(x1, y1, x2, y2, x3, y3, x4, y4)) //存在相交
        {
            add = getNum(max(x1, x3), max(y1, y3), min(x2, x4), min(y2, y4));
        }
        else
        {
            add.wnum = add.bnum = 0;
        }

        ans.wnum += first.bnum;
        ans.bnum -= first.bnum;

        /*相交部分：白→白→黑
                   黑→白→黑*/
        ans.wnum -= (add.wnum + add.bnum);
        ans.bnum += (add.wnum + add.bnum);

        ans.wnum -= (second.wnum - add.wnum);
        ans.bnum += (second.wnum - add.wnum);
        cout << ans.wnum << " " << ans.bnum << endl;
    }
    return 0;
}