挺好的一道题,一开始用lazy标记往下推,总是推不出样例的正解,然后就去看了相关博客,发现却确实如此,在这里是无法用lazy标记来层层推的,并且还会出现超内存的情况,所以,便改用了永久化标记来解这道题。
还有一件是,关于discuss里的讨论区有一块这样的讨论:
1
2 10
C 2 1 2 2
Q 2 3
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
相信很多人是过不了上面的数据的吧!——但是你们就没发现数据本身就错了吗。。。它开的N*N的空间不对啊,怎能询问空间外的点呢?应改为:
正确样例
1
3 10
C 2 1 2 2
Q 2 3
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
那么,我相信大家就都能过了吧。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define efs 1e-6
#define pi 3.141592653589793
#define e 2.718281828459045
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 1005;
int N, Q;
short int tree[maxN<<2][maxN<<2];
void pushup(int rt, int X0)
{
tree[X0][rt] = tree[X0][rt<<1] + tree[X0][rt<<1|1];
}
void update_In(int rt, int X0, int l, int r, int ql, int qr)
{
if(ql == l && qr == r)
{
tree[X0][rt]^=1;
return;
}
int mid = (l + r)>>1;
if(ql>mid) update_In(rt<<1|1, X0, mid+1, r, ql, qr);
else if(qr<=mid) update_In(rt<<1, X0, l, mid, ql, qr);
else
{
update_In(rt<<1, X0, l, mid, ql, mid);
update_In(rt<<1|1, X0, mid+1, r, mid+1, qr);
}
}
void update_Out(int rt, int l, int r, int qlx, int qly, int qrx, int qry)
{
if(qlx == l && qrx == r)
{
update_In(1, rt, 1, N, qly, qry);
return;
}
int mid = (l + r)>>1;
if(qlx>mid) update_Out(rt<<1|1, mid+1, r, qlx, qly, qrx, qry);
else if(qrx<=mid) update_Out(rt<<1, l, mid, qlx, qly, qrx, qry);
else
{
update_Out(rt<<1, l, mid, qlx, qly, mid, qry);
update_Out(rt<<1|1, mid+1, r, mid+1, qly, qrx, qry);
}
}
int query_In(int rt, int X0, int l, int r, int qy, int sum)
{
sum^=tree[X0][rt];
if(l == r) return sum;
int mid = (l + r)>>1;
if(qy>mid) return query_In(rt<<1|1, X0, mid+1, r, qy, sum);
else return query_In(rt<<1, X0, l, mid, qy, sum);
}
int query_Out(int rt, int l, int r, int qx, int qy, int sum)
{
sum^=query_In(1, rt, 1, N, qy, 0);
if(l == r) return sum;
int mid = (l + r)>>1;
if(qx>mid) return query_Out(rt<<1|1, mid+1, r, qx, qy, sum);
else return query_Out(rt<<1, l, mid, qx, qy, sum);
}
int main()
{
int T; scanf("%d", &T);
int Cas = 0;
while(T--)
{
if(Cas++ > 0) printf("\n");
scanf("%d%d", &N, &Q);
memset(tree, 0, sizeof(tree));
while(Q--)
{
char s[3];
scanf("%s", s);
if(s[0] == 'C')
{
int lx, ly, rx, ry;
scanf("%d%d%d%d", &lx, &ly, &rx, &ry);
update_Out(1, 1, N, lx, ly, rx, ry);
}
else
{
int x, y;
scanf("%d%d", &x, &y);
printf("%d\n", query_Out(1, 1, N, x, y, 0));
}
}
}
return 0;
}