CodeForce Div 2 C. Masha and two friends

题目链接： http://codeforces.com/contest/1080/problem/C

思路：双向延长两个矩形方块的4边，会形成一个被分割为9块的更大立方体。

计算所有的9个方框。方框中心的点只有可能在黑框，白框或者在外面。中心点在黑框中则按照黑色渲染了棋盘计算，其余同理。

详细见代码。

注意点：

1.题目上方框的信息给出的是方块号的信息而非坐标信息。需要转换成坐标信息

2.求两者的中点得用double转换，如果用float转换精度会有偏差。

3.sort函数应该是sort(x+1, x+5)，心态崩了。

 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <cstdio>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <stack>
 #define ll long long
 //#define local

 using namespace std;

 const int MOD = 1e9+;
 const int inf = 0x3f3f3f3f;
 const double PI = acos(-1.0);
 const int maxn = 3e5 + ;
 ll x1, x2, x3, x4;
 ll y11, y2, y3, y4;

 int judge(ll lowx, ll lowy, ll highx, ll highy) {
     double x = double(lowx + highx)/2.0;
     double y = double(lowy + highy)/2.0;
     if (x>x3 && x < x4 && y > y3 && y < y4) {
         return ;
     } else if (x>x1 && x < x2 && y > y11 && y < y2) {
         return ;
     } else {
         return ;
     }
 }

 int main() {
 #ifdef local
     if(freopen("/Users/Andrew/Desktop/data.txt", "r", stdin) == NULL) printf("can't open this file!\n");
 #endif

     int q;
     scanf("%d", &q);
     ll n, m;
     ll x[], y[];
     while (q--) {
         scanf("%lld%lld", &n, &m);
         ll white = n*m/;
         ll black = n*m/;
         if ((n*m) % ) {
             white++;
         }
         scanf("%lld%lld%lld%lld", &x1, &y11, &x2, &y2);
         scanf("%lld%lld%lld%lld", &x3, &y3, &x4, &y4);
         x1 -= ; x3 -= ;
         y11 -= ; y3 -= ;
         x[] = x1; x[] = x2;
         x[] = x3; x[] = x4;
         sort(x+,x+);
         y[] = y11; y[] = y2;
         y[] = y3; y[] = y4;
         sort(y+, y+);
         for (int i = ; i < ; ++i) {
             for (int j = ; j < ; ++j) {
                 if (x[i+]==x[i] || y[j+]==y[j]) continue;
                 ll tmp = (x[i+]-x[i])*(y[j+]-y[j]);
                 if (judge(x[i], y[j], x[i+], y[j+]) == ) {
                     continue;
                 } else if (judge(x[i], y[j], x[i+], y[j+]) == ) {
                     if (tmp % ) {
                         if ((x[i] + y[j]) %  == ) {// white
                             white += tmp/;
                             black -= tmp/;
                         }
                         else {
                             white += tmp/ + ;
                             black -= (tmp/+);
                         }
                     } else {
                         white += tmp/;
                         black -= tmp/;
                     }
                 } else {
                     if (tmp % ) {
                         if ((x[i] + y[j]) %  == ) {// black
                             white -= tmp/;
                             black += tmp/;
                         }
                         else {
                             white -= (tmp/+);
                             black += (tmp/+);
                         }
                     } else {
                         white -= tmp/;
                         black += tmp/;
                     }
                 }
             }
         }
         printf("%lld %lld\n", white, black);
     }

 #ifdef local
     fclose(stdin);
 #endif
     return ;
 }

思路2:

可以证明得到：白色与黑色相交方块的左下角坐标是:(max(x1, x3), max(y1, y3)) 右下角坐标是（min（x2, x4）, min(y2, y4)）