Rectangles
Time Limit: 5000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1259 Accepted Submission(s): 661
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.
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.
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
没优化版:
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 #include <math.h> 5 #include <algorithm> 6 using namespace std; 7 #define INF 100000000 8 typedef struct point 9 { 10 int x1,y1,x2,y2; 11 }point; 12 point p[100]; 13 int ans[1100000]={0}; 14 int n; 15 void dfs(int x1,int y1,int x2,int y2,int deep,int sign,int sta) 16 { 17 if( x1 >= x2 || y1 >= y2 ) return; 18 if(deep==n) 19 { 20 if(sta) 21 for(int i=1;i<(1<<n);i++) 22 { 23 if((i|sta)<=i) 24 ans[i]+=sign*(x2-x1)*(y2-y1); 25 } 26 return ; 27 } 28 dfs(x1,y1,x2,y2,deep+1,sign,sta); 29 dfs(max(x1,p[deep].x1),max(y1,p[deep].y1),min(x2,p[deep].x2),min(y2,p[deep].y2),deep+1,-sign,sta|(1<<deep)); 30 } 31 int main() 32 { 33 int m,i,ss,cas=1,mm,x,cass; 34 while(scanf("%d%d",&n,&m),(n||m)) 35 { 36 memset(ans,0,sizeof(ans)); 37 for(i=0;i<n;i++) 38 scanf("%d%d%d%d",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2); 39 dfs(0,0,INF,INF,0,-1,0); 40 printf("Case %d: ",cas++); 41 cass=1; 42 while(m--) 43 { 44 scanf("%d",&mm); 45 ss=0; 46 for(i=0;i<mm;i++) 47 { 48 scanf("%d",&x); 49 ss|=(1<<(x-1)); 50 } 51 printf("Query %d: %d ",cass++,ans[ss]); 52 } 53 printf(" "); 54 } 55 }
优化版:
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 #include <math.h> 5 #include <algorithm> 6 using namespace std; 7 #define INF 100000000 8 typedef struct point 9 { 10 int x1,y1,x2,y2; 11 } point; 12 point p[100]; 13 int ans[1100000],staa[100005]; 14 int n,m; 15 void dfs(int x1,int y1,int x2,int y2,int deep,int sign,int sta) 16 { 17 if( x1 >= x2 || y1 >= y2 ) return; 18 if(deep==n) 19 { 20 if(sta) 21 for(int i=0; i<m; i++) 22 { 23 if((staa[i]|sta)<=staa[i]) 24 ans[staa[i]]+=sign*(x2-x1)*(y2-y1); 25 } 26 return ; 27 } 28 dfs(x1,y1,x2,y2,deep+1,sign,sta); 29 dfs(max(x1,p[deep].x1),max(y1,p[deep].y1),min(x2,p[deep].x2),min(y2,p[deep].y2),deep+1,-sign,sta|(1<<deep)); 30 } 31 int main() 32 { 33 int i,cas=1,mm,x,cass; 34 while(scanf("%d%d",&n,&m),(n||m)) 35 { 36 memset(ans,0,sizeof(ans)); 37 memset(staa,0,sizeof(staa)); 38 for(i=0; i<n; i++) 39 scanf("%d%d%d%d",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2); 40 printf("Case %d: ",cas++); 41 cass=0; 42 while(m--) 43 { 44 scanf("%d",&mm); 45 for(i=0; i<mm; i++) 46 { 47 scanf("%d",&x); 48 staa[cass]|=(1<<(x-1)); 49 } 50 cass++; 51 } 52 m=cass; 53 dfs(0,0,INF,INF,0,-1,0); 54 for(i=1; i<=cass; i++) 55 printf("Query %d: %d ",i,ans[staa[i-1]]); 56 printf(" "); 57 } 58 }