poj2155 Matrix 树状数组套树状数组

Description

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给定一些二维数组中的操作形如C x1 y1 x2 y2表示把(x1,y1)到(x2,y2)中的数字0变成1，1变成0，Q x y表示查询(x,y)

Solution

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sb题，直接二维树状数组套树状数组。至于0还是1的问题可以看成操作了x次就是x%2，按照类似矩阵前缀和的方法更新

Code

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#include <stdio.h>#include <string.h>#define rep(i,st,ed) for (int i=st;i<=ed;++i)#define fill(x,t) memset(x,t,sizeof(x))#define N 2005int c[N][N];int T,n,m;void read(int &x) {    x=0; char ch=getchar();    for (;ch<'0'||ch>'9';ch=getchar());    for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());}int lowbit(int x) {return x&(-x);}int query(int x,int y) {    int ret=0;    for (int i=x;i>0;i-=lowbit(i)) {        for (int j=y;j>0;j-=lowbit(j)) {            ret+=c[i][j];        }    }    return ret&1;}void modify(int x,int y) {    for (int i=x;i<=n;i+=lowbit(i)) {        for (int j=y;j<=n;j+=lowbit(j)) {            ++c[i][j];        }    }}int main(void) {    read(T);    while (T--) {        fill(c,0);        read(n); read(m);        rep(i,1,m) {            char opt=getchar();            int x1,y1,x2,y2;            read(x1); read(y1);            if (opt=='Q') printf("%d\n",query(x1,y1));            else {                read(x2); read(y2);                modify(x1,y1);                modify(x2+1,y1);                modify(x1,y2+1);                modify(x2+1,y2+1);            }        }        puts("");    }    return 0;}