Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 3672 | Accepted: 1052 |
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
题意:求给定的R个矩形的面积的并。
思路:容斥原理,加上奇数的矩阵的面积的交,减去偶数的矩阵的面积的交。数据比较大,容易TLE。
AC代码如下:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct rec
{
int x1,y1,x2,y2,area;
}c[30],s[30];
int t,n,m,k,ans;
rec mul(rec a,rec b)
{
a.x1=max(a.x1,b.x1);a.y1=max(a.y1,b.y1);
a.x2=min(a.x2,b.x2);a.y2=min(a.y2,b.y2);
if(a.x1>=a.x2 || a.y1>=a.y2)
a.area=0;
else
a.area=(a.x2-a.x1)*(a.y2-a.y1);
return a;
}
void get_sum(rec a,int id,int num)
{
if(num&1)
ans+=a.area;
else
ans-=a.area;
int i,j;
for(i=id;i<=k;i++)
{
rec b=mul(a,s[i]);
if(b.area<=0)
continue;
get_sum(b,i+1,num+1);
}
}
void scan( int& x )
{
char c;
while( c = getchar(), c < '0' || c > '9' );
x = c - '0';
while( c = getchar(), c >= '0' && c <= '9' ) x = x * 10 + c - '0';
}
int main()
{
int i,j,k2;
while(~scanf("%d%d",&n,&m) &&(n!=0 || m!=0))
{
for(i=1;i<=n;i++)
{
scan(c[i].x1);scan(c[i].y1);scan(c[i].x2);scan(c[i].y2);
c[i].area=(c[i].x2-c[i].x1)*(c[i].y2-c[i].y1);
}
printf("Case %d:\n",++t);
for(i=1;i<=m;i++)
{
scan(k);
for(j=1;j<=k;j++)
{
scan(k2);
s[j]=c[k2];
}
ans=0;
for(j=1;j<=k;j++)
get_sum(s[j],j+1,1);
printf("Query %d: %d\n",i,ans);
}
printf("\n");
}
}