[容斥原理] hdu 2461 Rectangles

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=2461

Rectangles

Time Limit: 5000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1268    Accepted Submission(s): 665

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  20 0 2 21 1 3 31 12 1 22 10 1 1 22 1 3 22 1 20 0

 

Sample Output

Case 1:Query 1: 4Query 2: 7Case 2:Query 1: 2

 

Source

2008 Asia Hefei Regional Contest Online by USTC

 

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题目意思：

给n(n<=20)个矩形,有m(m<=100000)个询问，每个询问求给定矩形的并的面积。

解题思路：

简单容斥原理

对于指定的矩形并面积，先求出所有单个矩形的和，再减去两个矩形的交，再加上三个矩形的交，以此类推。

问题就转化为求多个矩形的交，这个转化为求两个矩形的交。

代码：

//#include<CSpreadSheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 33struct Rec{    int x1,y1,x2,y2;    bool vv;}rec[Maxn];int n,m,num,pp[Maxn],ans;Rec cal(Rec a,Rec b) //求两个矩形的交的矩形{    Rec res;    if(a.x1>b.x1)        swap(a,b);    if(b.x1>=a.x2||b.y1>=a.y2||b.y2<=a.y1)    {        res.vv=false;  //说明交的面积为0        return res;    }    res.vv=true; //注意要置为true    res.x1=b.x1;    res.y1=max(a.y1,b.y1);    res.x2=min(a.x2,b.x2);    res.y2=min(a.y2,b.y2);    return res;}int area(Rec cur){    return (cur.x2-cur.x1)*(cur.y2-cur.y1);}void dfs(Rec hav,int cur,int cn){    if(cur>num)        return ;    for(int i=cur;i<=num;i++)    {        Rec temp=cal(hav,rec[pp[i]]);        if(!temp.vv)  //没有公共部分            continue;        if(cn&1)            ans+=area(temp); //求公共部分面积        else            ans-=area(temp);        dfs(temp,i+1,cn^1);    }}int main(){   //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   int ca=0;   while(scanf("%d%d",&n,&m)&&n+m)   {       printf("Case %d:\n",++ca);       for(int i=1;i<=n;i++)           scanf("%d%d%d%d",&rec[i].x1,&rec[i].y1,&rec[i].x2,&rec[i].y2);       for(int qu=1;qu<=m;qu++)       {           ans=0;           scanf("%d",&num);           for(int i=1;i<=num;i++)               scanf("%d",&pp[i]);           for(int i=1;i<=num;i++)           {               ans+=area(rec[pp[i]]);               dfs(rec[pp[i]],i+1,0);           }           printf("Query %d: %d\n",qu,ans);       }       putchar('\n');   }   return 0;}