POJ2155 Matrix
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 25843 | Accepted: 9545 |
Description
Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N).
We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.
1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].
We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.
1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].
Input
The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.
The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.
The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.
Output
For each querying output one line, which has an integer representing A[x, y].
There is a blank line between every two continuous test cases.
There is a blank line between every two continuous test cases.
Sample Input
1 2 10 C 2 1 2 2 Q 2 2 C 2 1 2 1 Q 1 1 C 1 1 2 1 C 1 2 1 2 C 1 1 2 2 Q 1 1 C 1 1 2 1 Q 2 1
Sample Output
1 0 0 1
Source
POJ Monthly,Lou Tiancheng
二维线段树
区间修改,单点查询。
修改时,区间内0改成1,1改成0。这里等价成每次值+1,求解时%2
二维线段树的标记无法下传,解决方法是标记永久化,求值时,累积树上每一层的标记。
RE了好久,发现是边界写错了……
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 #define ls l,mid,rt<<1 6 #define rs mid+1,r,rt<<1|1 7 using namespace std; 8 const int mxn=4020; 9 int t[mxn][mxn]; 10 int n,m; 11 int read(){ 12 int x=0,f=1;char ch=getchar(); 13 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 14 while(ch>='0' && ch<='9'){x=x*10-'0'+ch;ch=getchar();} 15 return x*f; 16 } 17 void updateY(int L,int R,int v,int l,int r,int rt,int px){ 18 // printf("UY:L:%d R:%d v:%d %d %d %d %d\n",L,R,v,l,r,rt,px); 19 if(L<=l && r<=R){ 20 t[px][rt]+=v;return; 21 } 22 int mid=(l+r)>>1; 23 if(L<=mid)updateY(L,R,v,ls,px); 24 if(R>mid)updateY(L,R,v,rs,px); 25 return; 26 } 27 void updateX(int L,int R,int yy1,int yy2,int v,int l,int r,int rt){ 28 if(L<=l && r<=R){ 29 updateY(yy1,yy2,v,1,n,1,rt);return; 30 } 31 int mid=(l+r)>>1; 32 if(L<=mid)updateX(L,R,yy1,yy2,v,ls); 33 if(R>mid)updateX(L,R,yy1,yy2,v,rs); 34 return; 35 } 36 int qY(int y,int l,int r,int rt,int px){ 37 if(l==r){return t[px][rt];} 38 int mid=(l+r)>>1; 39 if(y<=mid)return t[px][rt]+qY(y,ls,px); 40 else return t[px][rt]+qY(y,rs,px); 41 } 42 int qX(int x,int y,int l,int r,int rt){ 43 int res=qY(y,1,n,1,rt); 44 if(l==r)return res; 45 int mid=(l+r)>>1; 46 if(x<=mid)return res+qX(x,y,ls); 47 else return res+qX(x,y,rs); 48 } 49 int main(){ 50 int T=read(); 51 int i,j; 52 int x1,x2,yy1,yy2; 53 while(T--){ 54 memset(t,0,sizeof t); 55 // 56 n=read();m=read(); 57 char op[3]; 58 while(m--){ 59 scanf("%s",op); 60 if(op[0]=='C'){ 61 x1=read(),yy1=read(),x2=read(),yy2=read(); 62 updateX(x1,x2,yy1,yy2,1,1,n,1); 63 } 64 else{ 65 int x1=read(),yy1=read(); 66 int ans=qX(x1,yy1,1,n,1); 67 if(ans&1)printf("1\n"); 68 else printf("0\n"); 69 } 70 } 71 printf("\n"); 72 } 73 return 0; 74 }
本文为博主原创文章,转载请注明出处。