2015
02-22

There are no more than 10 test cases
For each case, the first line contains four integers C , N , M,S ( 1<= C <= 20 ,1 ≤ N,M <= 1000, 1<= S <= 50 ,M%2==0), indicating the number of queries , the number of rectangles, the number of images, the size of image .Noting ：all the images are square and the same in size .
Then there are N lines. Each line has four integers x1 y1 x2 y2 (x1 <= x2, y1 <= y2, 0 <= x1, x2, y1, y2 < S) .They are the coordinates of the left-top and the right-bottom points of the rectangle . (y is row ,x is column)
Then there are M lines. Each lines has S*S integers. All integers from 0 to 255.
The ith Image is face when i is between 1 and M/2 , notface otherwise.
Then there are C lines. Each line has M integers. The ith integer is qi indicating to weight of ith image.( 1<= qi <= 2000000 )

There are no more than 10 test cases
For each case, the first line contains four integers C , N , M,S ( 1<= C <= 20 ,1 ≤ N,M <= 1000, 1<= S <= 50 ,M%2==0), indicating the number of queries , the number of rectangles, the number of images, the size of image .Noting ：all the images are square and the same in size .
Then there are N lines. Each line has four integers x1 y1 x2 y2 (x1 <= x2, y1 <= y2, 0 <= x1, x2, y1, y2 < S) .They are the coordinates of the left-top and the right-bottom points of the rectangle . (y is row ,x is column)
Then there are M lines. Each lines has S*S integers. All integers from 0 to 255.
The ith Image is face when i is between 1 and M/2 , notface otherwise.
Then there are C lines. Each line has M integers. The ith integer is qi indicating to weight of ith image.( 1<= qi <= 2000000 )

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

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

Case #1:
1
Case #2:
0

HintThe first sample f1 (x1) = 1, f1 (x2) = 4, f1 (x3) = 2, f1 (x4 ) = 3, so when θ ∈ (1, 2] , p = 1, or θ ∈ [3, 4), p = -1, ∈t = 1.


1. 其实国内大部分公司对算法都不够重视。特别是中小型公司老板根本都不懂技术，也不懂什么是算法，从而也不要求程序员懂什么算法，做程序从来不考虑性能问题，只要页面能显示出来就是好程序，这是国内的现状，很无奈。

2. 很高兴你会喜欢这个网站。目前还没有一个开发团队，网站是我一个人在维护，都是用的开源系统，也没有太多需要开发的部分，主要是内容整理。非常感谢你的关注。

3. 在方法1里面：

//遍历所有的边，计算入度
for(int i=0; i<V; i++)
{
degree = 0;