二维树状数组-POJ-2155-Matrix
2016-05-10 18:03
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Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 23707 Accepted: 8762
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.
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).
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.
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.
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
题意:
给你一个n*n的初始全部为0的矩阵,然后进行t次操作,每次可能有两种操作,一种是让矩阵中一部分全部取非运算,其中(x1,y1)是最左上的点,(x2,y2)是最右下的点;另一种是查询(x,y)的值。
题解:
这是楼教主当年出的题,是经典的二维树状数组,区间修改单点查询的题,注意处理Change时几个矩形的加减法就好。
在这道题做了个测试,关了同步的cin cout还是比scanf printf慢了不少。
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 23707 Accepted: 8762
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.
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).
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.
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.
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
题意:
给你一个n*n的初始全部为0的矩阵,然后进行t次操作,每次可能有两种操作,一种是让矩阵中一部分全部取非运算,其中(x1,y1)是最左上的点,(x2,y2)是最右下的点;另一种是查询(x,y)的值。
题解:
这是楼教主当年出的题,是经典的二维树状数组,区间修改单点查询的题,注意处理Change时几个矩形的加减法就好。
在这道题做了个测试,关了同步的cin cout还是比scanf printf慢了不少。
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <queue> #include <vector> #include <set> #include <utility> #define LL long long int using namespace std; const int MAXN=1005; int T,n; LL t,a[MAXN][MAXN]; inline LL LowBit(LL num) { return num&(-num); } inline void Change(int x,int y,int num) { if(x<=0 || y<=0) return; for(int i=x;i>0;i-=LowBit(i)) for(int j=y;j>0;j-=LowBit(j)) a[i][j]+=num; } inline int GetVaule(int x,int y) { LL sum=0; for(int i=x;i<=n;i+=LowBit(i)) for(int j=y;j<=n;j+=LowBit(j)) sum+=a[i][j]; return sum%2; } int main(void) { char tmp; int x1,x2,y1,y2; scanf("%lld",&T); while(T--) { memset(a,0,sizeof(a)); scanf("%d%lld",&n,&t); for(LL i=1;i<=t;i++) { cin >> tmp; switch(tmp){ case 'C': scanf("%d%d%d%d",&x1,&y1,&x2,&y2); Change(x2,y2,1);Change(x1-1,y1-1,1);Change(x2,y1-1,-1);Change(x1-1,y2,-1); break; case 'Q':scanf("%d%d",&x1,&y1); printf("%d\n",GetVaule(x1,y1)); } } if(T) printf("\n"); } return 0; }
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