Hdu2461 Rectangles ----- reject

Rectangles
Time Limit: 5000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 989 Accepted Submission (s): 558

Problem Description
You are developing a software for painting rectangles on the screen. the software supports drawing several rectangles and filling some of them with a color different from the color of the background. you are to implement an important function. the function answer such queries as what is the colored area if a subset of rectangles on the screen are filled.

Input
The input consists of multiple test cases. each test case starts with a line containing two integers N (1 ≤ N ≤ 20) and M (1 ≤m ≤ 100000 ), indicating the number of rectangles on the screen and the number of queries, respectively.
The I-th line of the following N lines contains four integers X1, Y1, X2, Y2 (0 ≤ X1 <X2 ≤1000, 0 ≤ Y1 <Y2 ≤ 1000 ), which indicate that the lower-left and upper-right coordinates of the I-th rectangle are (X1, Y1) and (X2, Y2 ). rectangles are numbered from 1 to N.
The last M lines of each test case describe M queries. each query starts with a integer R (1 <= R ≤ N), which is the number of rectangles the query is supposed to fill. the following list of R integers in the same line gives the rectangles the query is supposed to fill, each integer of which will be between 1 and N, inclusive.

The last test case is followed by a line containing two zeros.

Output
For each test case, print a line containing the test case number (beginning with 1 ).
For each query in the input, print a line containing the query number (beginning with 1) followed by the corresponding answer for the query. Print a blank line after the output for each test case.

Sample Input
2 2
0 0 2 2
1 1 3 3
1 1
2 1 2
2 1
0 1 1 2
2 1 3 2
2 1 2
0 0

Sample Output
Case 1:
Query 1: 4
Query 2: 7

Case 2:
Query 1: 2

Source
2008 Asia Hefei Regional Contest Online by USTC

Recommend
Teddy
Question: give n rectangles, then apply m rectangles each time, and ask how much area the colored area is.
You can use the refresh principle.
[Cpp]
# Include <iostream>
# Include <cstdlib>
# Include <stdio. h>
# Include <algorithm>
Using namespace std;
Int n, m, num, res;
Int a [25];
Struct Node
{
Int x1, y1, x2, y2;
Int s;
} Node [25];
Bool check (Node x, int u)
{
If (node [u]. x1> = x. x2) return 0;
If (x. x1> = node [u]. x2) return 0;
If (node [u]. y1> = x. y2) return 0;
If (x. y1> = node [u]. y2) return 0;
Return 1;
}
Void solve (Node rec, int id, int k)
{
Node temp;
If (id = num + 1) return;
For (int I = id; I <= num; I ++)
{
If (check (rec, a [I])
{
Temp. x1 = rec. x1> node [a [I]. x1? Rec. x1: node [a [I]. x1;
Temp. y1 = rec. y1> node [a [I]. y1? Rec. y1: node [a [I]. y1;
Temp. x2 = rec. x2 <node [a [I]. x2? Rec. x2: node [a [I]. x2;
Temp. y2 = rec. y2 <node [a [I]. y2? Rec. y2: node [a [I]. y2;
Temp. s = (temp. x2-temp.x1) * (temp. y2-temp.y1 );
If (k % 2) res + = temp. s;
Else res-= temp. s;
Solve (temp, I + 1, k + 1 );
}
}
}
Int main ()
{
Int count = 1;
While (scanf ("% d", & n, & m )! = EOF)
{
If (n = 0 & m = 0) break;
For (int I = 1; I <= n; I ++)
{
Scanf ("% d", & node [I]. x1, & node [I]. y1, & node [I]. x2, & node [I]. y2 );
Node [I]. s = (node [I]. x2-node [I]. x1) * (node [I]. y2-node [I]. y1 );
}
Printf ("Case % d: \ n", count ++ );
For (int I = 1; I <= m; I ++)
{
Scanf ("% d", & num );
Res = 0;
For (int j = 1; j <= num; j ++)
{
Scanf ("% d", & a [j]);
Res + = node [a [j]. s;
}
For (int j = 1; j <= num; j ++)
{
Solve (node [a [j], j +, 0 );
}
Printf ("Query % d:", I );
Printf ("% d \ n", res );
}
Puts ("");
}
Return 0;
}