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poj2155 Matrix 二維數組數組

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poj2155 Matrix 二維數組數組

http://poj.org/problem?id=2155

Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 18769 Accepted: 7078

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

Sample Output

Source

POJ Monthly,Lou Tiancheng
題意：二維矩陣，可以把子矩陣反轉，查詢某點的狀態。題解：二維樹狀數組可搞。屬於樹狀數組的第二類應用，區間修改單點查值。

（2）“改段求點”型，即對於序列A有以下操作：

【1】修改操作：將A[l..r]之間的全部元素值加上c；

【2】求和操作：求此時A[x]的值。

這個模型中需要設置一個輔助數組B：B[i]表示A[1..i]到目前為止共被整體加了多少（或者可以說成，到目前為止的所有ADD(i, c)操作中c的總和）。

則可以發現，對於之前的所有ADD(x, c)操作，當且僅當x>=i時，該操作會對A[i]的值造成影響（將A[i]加上c），又由於初始A[i]=0，所以有A[i] = B[i..N]之和！而ADD(i, c)（將A[1..i]整體加上c），將B[i]加上c即可——只要對B數組進行操作就行了。

這樣就把該模型轉化成了“改點求段”型，只是有一點不同的是，SUM(x)不是求B[1..x]的和而是求B[x..N]的和，此時只需把ADD和SUM中的增減次序對調即可（模型1中是ADD加SUM減，這裡是ADD減SUM加）。代碼：

void ADD(int x, int c)
{
     for (int i=x; i>0; i-=i&(-i)) b[i] += c;
}
int SUM(int x)
{
    int s = 0;
    for (int i=x; i<=n; i+=i&(-i)) s += b[i];
    return s;
}

操作【1】：ADD(l-1, -c); ADD(r, c);

操作【2】：SUM(x)。

對於這道題，就是變為二維的而已，取反操作可以看成+1,最後就看是否整除2即可。比如操作x1,y1,x2,y2的子矩形，就是 add(x2,y2,1);
add(x1-1,y2,-1);
add(x2,y1-1,-1);
add(x1-1,y1-1,1);
畫個圖想想就知道了。代碼：

/**
 * @author neko01
 */
//#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
const LL inf=(((LL)1)<<61)+5;
const int N=1005;
int bit[N][N];
int n;
int sum(int x,int y)
{
    int s=0;
    for(int i=x;i<=n;i+=i&-i)
        for(int j=y;j<=n;j+=j&-j)
            s+=bit[i][j];
    return s;
}
void add(int x,int y,int val)
{
    for(int i=x;i>0;i-=i&-i)
        for(int j=y;j>0;j-=j&-j)
            bit[i][j]+=val;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int q;
        scanf("%d%d",&n,&q);
        clr(bit);
        while(q--)
        {
            char s[5];
            scanf("%s",s);
            int x1,x2,y1,y2;
            if(s[0]=='C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                add(x2,y2,1);
                add(x1-1,y2,-1);
                add(x2,y1-1,-1);
                add(x1-1,y1-1,1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                printf("%d\n",sum(x1,y1)%2);
            }
        }
        puts("");
    }
    return 0;
}