poj 2155 Matrix

Matrix

Time Limit: 3000MS		Memory Limit: 65536K

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

1
0
0
1
题目大意：
n*n矩阵
一次访问一个子矩阵
一次询问一个点的访问次数%2
二维线段树

#include<cstdio>
#include<cstring>
#define N 1001
using namespace std;
int T,n,m,ans,f[N*4][N*4];
char c[2];int x1,x2,y1,y2;
void changey(int kx,int ky,int ly,int ry)
{
    if(y1<=ly&&ry<=y2)
    {
        f[kx][ky]++;
        return;
    }
    int mid=ly+ry>>1;
    if(y1<=mid) changey(kx,ky<<1,ly,mid);
    if(y2>mid) changey(kx,(ky<<1)+1,mid+1,ry);
}
void changex(int kx,int lx,int rx)
{
    if(x1<=lx&&rx<=x2)
    {
        changey(kx,1,1,n);
        return;
    }
    int mid=lx+rx>>1;
    if(x1<=mid) changex(kx<<1,lx,mid);
    if(x2>mid) changex((kx<<1)+1,mid+1,rx);
}
void queryy(int kx,int ky,int ly,int ry)
{
    ans+=f[kx][ky];
    if(ly==ry) return;
    int mid=ly+ry>>1;
    if(y1<=mid) queryy(kx,ky<<1,ly,mid);
    else queryy(kx,(ky<<1)+1,mid+1,ry);
}
void queryx(int kx,int lx,int rx)
{
    queryy(kx,1,1,n);
    if(lx==rx) return;
    int mid=lx+rx>>1;
    if(x1<=mid) queryx(kx<<1,lx,mid);
    else queryx((kx<<1)+1,mid+1,rx);
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        memset(f,0,sizeof(f));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%s",c);
            if(c[0]=='C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                changex(1,1,n);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                ans=0;
                queryx(1,1,n);
                printf("%d
",ans%2);
            }        
        }
        printf("
");
    }    
}

View Code

二维树状数组

#include<cstdio>
#include<cstring>
#define N 1001
using namespace std;
int T,n,m,tot,ans;
int f[N][N];
int lowbit(int x)
{
    return x&(-x);
}
void change(int x,int y)
{
    for(int i=x;i<=n;i+=lowbit(i))
     for(int j=y;j<=n;j+=lowbit(j))
      f[i][j]++;
} 
void query(int x,int y)
{
    for(int i=x;i;i-=lowbit(i))
     for(int j=y;j;j-=lowbit(j))
      ans+=f[i][j];
}
int main()
{
    scanf("%d",&T);int e=0;
    char c[2];int x1,x2,y1,y2;
    while(T--)
    {
        memset(f,0,sizeof(f));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%s",c);
            if(c[0]=='C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                change(x1,y1);
                change(x1,y2+1);
                change(x2+1,y1);
                change(x2+1,y2+1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                ans=0;
                query(x1,y1);
                printf("%d
",ans%2);
            }        
        }
        printf("
");
    }    
}

View Code

超时的自己写的二维线段树

#include<cstdio>
#include<cstring>
#define N 1201
using namespace std;
int T,n,m,tot,ans;
struct node
{
    int sum,f;
    int lx,ly,rx,ry;
    int son[4];
    void clear() 
    {
        sum=f=0;
        memset(son,0,sizeof(son));
    }
}tr[16*N*N];
void build(int & k,int lx,int ly,int rx,int ry)
{
    k=++tot;
    if(rx<lx||ry<ly) return;
    tr[k].clear();
    tr[k].lx=lx;tr[k].ly=ly;tr[k].rx=rx;tr[k].ry=ry;    
    if(lx==rx&&ly==ry) return;
    int midx=lx+rx>>1,midy=ly+ry>>1;
    build(tr[k].son[0],lx,ly,midx,midy);
    build(tr[k].son[1],lx,midy+1,midx,ry);
    build(tr[k].son[2],midx+1,ly,rx,midy);
    build(tr[k].son[3],midx+1,midy+1,rx,ry);
}
void down(int k)
{
    if(tr[k].son[0]) tr[tr[k].son[0]].sum+=tr[k].f,tr[tr[k].son[0]].f+=tr[k].f;
    if(tr[k].son[1]) tr[tr[k].son[1]].sum+=tr[k].f,tr[tr[k].son[1]].f+=tr[k].f;
    if(tr[k].son[2]) tr[tr[k].son[2]].sum+=tr[k].f,tr[tr[k].son[2]].f+=tr[k].f;
    if(tr[k].son[3]) tr[tr[k].son[3]].sum+=tr[k].f,tr[tr[k].son[3]].f+=tr[k].f;
    tr[k].f=0;
}
void work(int k,int oplx,int oply,int oprx,int opry,int p)
{
    if(tr[k].lx>=oplx&&tr[k].rx<=oprx&&tr[k].ly>=oply&&tr[k].ry<=opry)
    {
        if(p==1)
        {
            tr[k].sum++;
            tr[k].f++;
        }
        else ans=tr[k].sum;
        return ;
    }
    if(tr[k].f) down(k);
    int midx=tr[k].lx+tr[k].rx>>1,midy=tr[k].ly+tr[k].ry>>1;
    if(oplx<=midx&&oply<=midy) work(tr[k].son[0],oplx,oply,oprx,opry,p);
    if(oplx<=midx&&opry>midy) work(tr[k].son[1],oplx,oply,oprx,opry,p);
    if(oprx>midx&&oply<=midy) work(tr[k].son[2],oplx,oply,oprx,opry,p);
    if(oprx>midx&&opry>midy) work(tr[k].son[3],oplx,oply,oprx,opry,p);
}
int main()
{
    scanf("%d",&T);int e=0;
    char c[2];int x1,x2,y1,y2;
    while(T--)
    {
        tot=0;
        scanf("%d%d",&n,&m);
        build(e,1,1,n,n);
        for(int i=1;i<=m;i++)
        {
            scanf("%s",c);
            if(c[0]=='C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                work(1,x1,y1,x2,y2,1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                work(1,x1,y1,x1,y1,2);
                printf("%d
",ans%2);
            }        
        }
    }    
}

View Code