扫描线
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：
输入n 个矩形，m 次询问，下面 n 行输入矩形的左下角和右上角坐标 ，下面m 次询问，每次第一个数 k 表示后面跟 k 个数，询问 这 k 个矩形的合并面积。
思路：
每次询问都对这 k 个矩形做一次标记，开始计算覆盖面积，只不过不用线段树。
对着 k 个矩形按照 纵坐标 y 从小到大排序，，离散化这些矩形的横坐标 ， 然后对于遍历横坐标，同时寻找每两个横坐标之间的最大最小值（连续的 ，对于非连续的直接把之前的U D计算相加）最后计算这一块矩形的面积。还是跟切割矩形差不多
撸代码：
#include<algorithm>
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
struct node
{
int x1,x2,y1,y2;
int l;
} p[30],q[50];
bool comp(node a,node b)
{
return a.y1<b.y1;
}
int book[30],x[1010];
int work(int n)
{
int cnt=0;
for(int i=1; i<=n; i++)
{
q[i]=p[book[i]];
x[cnt++]=p[book[i]].x1;
x[cnt++]=p[book[i]].x2;
}
sort(x,x+cnt);
sort(q+1,q+n+1,comp);
int m=1;
/**离散化处理*/
for(int i=1; i<cnt; i++)
{
if(x[i]!=x[i-1])
x[m++]=x[i];
}
int L,R,sum=0,U,D,INF=0x3f3f3f3f;/**U D 计算最大最小值*/
for(int i=1; i<m; i++)
{
L=x[i-1],R=x[i];
U=0,D=INF;
for(int j=1; j<=n; j++)
{
if(q[j].x1<=L&&q[j].x2>=R)
{
if(q[j].y1>U&&U>D)/**有断层的直接先把之前的加上，换U ，D*/
{
sum+=(R-L)*(U-D);
U=q[j].y2,D=q[j].y1;
}
else
{
U=max(U,q[j].y2);
D=min(D,q[j].y1);
}
}
}
if(U>D)
{
sum+=(R-L)*(U-D);
}
}
return sum;
}
int main()
{
int n,m;
int t=1;
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0)
break;
for(int i=1; i<=n; i++)
scanf("%d%d%d%d",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2);
printf("Case %d:
",t++);
for(int i=1,k,x; i<=m; i++)
{
scanf("%d",&k);
memset(book,0,sizeof(book));
for(int j=1; j<=k; j++)
{
scanf("%d",&book[j]);
}
printf("Query %d: %d
",i,work(k));
}
printf("
");
}
return 0;
}