【二維樹狀陣列】poj 2155 Matrix
阿新 • • 發佈:2020-11-25
Matrix
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].
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.
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.
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 38085 | Accepted: 13627 |
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].
Input
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
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
程式碼
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<map> using namespace std; const int maxn=1005; int t,cc[maxn][maxn],n; int lowbit(int x) { return x&(-x); } void add(int x,int y,int v) { for(int i=x;i<=n;i+=lowbit(i)) for(int j=y;j<=n;j+=lowbit(j)) cc[i][j]+=v; } int query(int x,int y) { int res=0; for(int i=x;i>0;i-=lowbit(i)) for(int j=y;j>0;j-=lowbit(j)) res+=cc[i][j]; return res; } int main() { freopen("a.in","r",stdin); freopen("a.out","w",stdout); scanf("%d",&t); char e; int a,b,c,d,m; while(t--) { scanf("%d%d",&n,&m); memset(cc,0,sizeof(cc)); for(int i=1;i<=m;i++) { getchar(); e=getchar(); if(e=='Q') { scanf("%d%d",&a,&b); a++; b++; int ans=query(a-1,b-1); printf("%d\n",ans%2); } if(e=='C') { scanf("%d%d%d%d",&a,&b,&c,&d); a++; b++; c++; d++; add(a-1,b-1,1); add(c,d,1); add(a-1,d,1); add(c,b-1,1); } } if(t) printf("\n"); } return 0; }