Rectangles(hdu2461)

Rectangles

Time Limit: 5000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1497    Accepted Submission(s): 773


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.
思路：容斥原理；
感觉这题的数据有点水，按照这个复杂度O（2^n*m）是 感觉会超时的；
跑了1200多Ms。
思路很简单，就是容斥求面积。
  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<queue>
  6 #include<stack>
  7 #include<map>
  8 using namespace std;
  9 typedef struct pp
 10 {
 11         int x1;
 12         int y1;
 13         int x2;
 14         int y2;
 15 } ss;
 16 ss ju[30];
 17 int quer[1<<22];
 18 int cnt[30];
 19 int aa[1<<22];
 20 int bt[30];
 21 int cp[100005];
 22 int ask[100005];
 23 int tt[100];
 24 int N;
 25 void dfs(int k,int u);
 26 int main(void)
 27 {
 28         int i,j,k,p,q;
 29         int kk=0;
 30         while(scanf("%d %d",&p,&q),p!=0&&q!=0)
 31         {
 32                 kk++;
 33                 memset(ask,0,sizeof(ask));
 34                 for(i=0; i<p; i++)
 35                 {
 36                         scanf("%d %d %d %d",&ju[i].x1,&ju[i].y1,&ju[i].x2,&ju[i].y2);
 37                 }
 38                 int s;
 39                 int n;
 40                 for(s=0; s<q; s++)
 41                 {
 42                         N=0;
 43                         scanf("%d",&n);
 44                         int mn=0;
 45                         int dd;
 46                         for(i=0; i<n; i++)
 47                         {
 48                                 scanf("%d",&dd);
 49                                 mn|=(1<<(dd-1));
 50                         }
 51                         cp[s]=mn;
 52                 }
 53                 int flag[23];
 54                 for(i=1; i<=(1<<p)-1; i++)
 55                 {
 56                         int cn=0;
 57                         memset(flag,0,sizeof(flag));
 58                         for(j=0; j<p; j++)
 59                         {
 60                                 if(i&(1<<j))
 61                                 {
 62                                         cn++;
 63                                         flag[j]=1;
 64                                 }
 65                         }
 66                         int nn;
 67                         int ak=0;
 68                         for(nn=0; nn<22; nn++)
 69                         {
 70                                 if(flag[nn])
 71                                 {
 72                                         cnt[ak++]=nn;
 73                                 }
 74                         }
 75                         int mm;
 76                         int xx1,yy1,xx2,yy2;
 77                         xx1=ju[cnt[0]].x1;
 78                         yy1=ju[cnt[0]].y1;
 79                         xx2=ju[cnt[0]].x2;
 80                         yy2=ju[cnt[0]].y2;
 81                         int uu=0;
 82                         for(nn=1; nn<ak; nn++)
 83                         {
 84                                 if(xx1>=ju[cnt[nn]].x2)
 85                                 {
 86                                         uu=1;
 87                                         aa[i]=0;
 88                                         break;
 89                                 }
 90                                 else if(xx2<=ju[cnt[nn]].x1)
 91                                 {
 92                                         uu=1;
 93                                         aa[i]=0;
 94                                         break;
 95                                 }
 96                                 else if(yy1>=ju[cnt[nn]].y2)
 97                                 {
 98                                         uu=1;
 99                                         aa[i]=0;
100                                         break;
101                                 }
102                                 else if(yy2<=ju[cnt[nn]].y1)
103                                 {
104                                         uu=1;
105                                         aa[i]=0;
106                                         break;
107                                 }
108                                 else
109                                 {
110                                         xx1=max(xx1,ju[cnt[nn]].x1);
111                                         yy1=max(yy1,ju[cnt[nn]].y1);
112                                         xx2=min(xx2,ju[cnt[nn]].x2);
113                                         yy2=min(yy2,ju[cnt[nn]].y2);
114                                 }
115                         }
116                         if(!uu)
117                         {
118                                 int miji=abs(xx1-xx2)*abs(yy1-yy2);
119                                 aa[i]=miji;
120                                 for(s=0; s<q; s++)
121                                 {
122                                         int gg=cp[s]|i;
123                                         if(gg<=cp[s])
124                                         {
125                                                 if(cn%2)
126                                                 {
127                                                         ask[s]+=aa[i];
128                                                 }
129                                                 else ask[s]-=aa[i];
130                                         }
131                                 }
132                         }
133                 }
134                 printf("Case %d:\n",kk);
135                 for(j=0; j<q; j++)
136                 {
137                         printf("Query %d: ",j+1);
138                         printf("%d\n",ask[j]);
139                 }
140                 printf("\n");
141         }
142         return 0;
143 }

 

 
posted @ 2016-05-05 18:18  sCjTyC  阅读(413)  评论(0)  编辑  收藏  举报