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二维树状数组-POJ-2155-Matrix

来源:互联网 作者:佚名 时间:2016-05-12 15:58
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]

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.

  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.

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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