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poj 2155 Matrix
Matrix
http://poj.org/problem?id=2155
| 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].
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.
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.
There is a blank line between every two continuous test cases.
Sample Input
12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1
Sample Output
1001
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 1001using 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\n",ans%2); } } printf("\n"); } }
二维树状数组
#include<cstdio>#include<cstring>#define N 1001using 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\n",ans%2); } } printf("\n"); } }
超时的自己写的二维线段树
#include<cstdio>#include<cstring>#define N 1201using 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\n",ans%2); } } } }
poj 2155 Matrix
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