CodeChef：Magic Board（思维 & 树状数组）

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Long long ago, there is a magic board. The magic board has N*N cells: N rows and Ncolumns. Every cell contains one integer, which is 0 initially. Let the rows and the columns be numbered from 1 to N.

There are 2 types of operations can be applied to the magic board:

1. RowSet i x: it means that all integers in the cells on row i have been changed to x after this operation.
2. ColSet i x: it means that all integers in the cells on column i have been changed to x after this operation.

And your girlfriend sometimes has an interest in the total number of the integers 0s on some row or column.

1. RowQuery i: it means that you should answer the total number of 0s on row i.
2. ColQuery i: it means that you should answer the total number of 0s on column i.

Input

The first line of input contains 2 space-separated integers N and Q. They indicate the size of the magic board, and the total number of operations and queries from the girlfriend.

Then each of the next Q lines contains an operation or a query by the format mentioned above.

Output

For each query, output the answer of the query.

Constraints

1 ≤ N, Q ≤ 500000 (5 * 10⁵)
1 ≤ i ≤ N
x ∈ {0, 1} (That is, x = 0 or 1)

Sample

Input:3 6RowQuery 1ColSet 1 1RowQuery 1ColQuery 1RowSet 1 0ColQuery 1Output:3201

Explanation

At first, the magic board is

000 <- row 1000000

So the answer of first query "RowQuery 1" is 3.

After the "ColSet 1 1", the board becomes

column 1|V100100100

So the answer of the second query "RowQuery 1" is clearly 2, since the cell (1,1)became 1. And the answer of the third query "ColQuery 1" is 0.

Finally, apply the operation "RowSet 1 0", the board has changed to

000100100

From this board, the answer of the last query "ColQuery 1" should be 1.

题意：Q个操作，修改操作为将一行或一列设置成x，查询操作为输出某一行或列中0的数量。

思路：加如要查询行i的0数量，先判断最后一次修改行i是什么值及时间t，如果是1，找出有几列是t之后被修改为0的，这些可以通过时间戳+Bit树完成。

# include <bits/stdc++.h># define mp make_pair# define pii pair<int,int># define A first# define B secondusing namespace std;const int maxn = 5e5+10;pii c[maxn+3], r[maxn+3];int c0[maxn+3], c1[maxn+3], r0[maxn+3], r1[maxn+3];inline void update(int a[], int pos, int val){    if(pos == 0) return;    for(int i=pos; i<=maxn; i+=i&(-i))        a[i] += val;}inline int query(int a[], int pos){    if(pos == 0) return 0;    int ans = 0;    for(int i=pos; i; i-=i&(-i))        ans += a[i];    return ans;}int main(){    int n, q, x, y;    char s[13];    scanf("%d%d",&n,&q);    for(int i=1; i<=q; ++i)    {        scanf("%s",s);        if(s[3] == 'S')        {            scanf("%d%d",&x,&y);            if(s[0] == 'R')            {                if(r[x].B) update(r1, r[x].A, -1);                else update(r0, r[x].A, -1);                if(y) update(r1, i, 1);                else update(r0, i, 1);                r[x] = mp(i, y);            }            else            {                if(c[x].B) update(c1, c[x].A, -1);                else update(c0, c[x].A, -1);                if(y) update(c1, i, 1);                else update(c0, i, 1);                c[x] = mp(i, y);            }        }        else        {            scanf("%d",&x);            int ans = 0;            if(s[0] == 'R')            {                if(r[x].B) ans = query(c0, i) - query(c0, r[x].A);                else ans = n - query(c1, i) + query(c1, r[x].A);            }            else            {                if(c[x].B) ans = query(r0, i) - query(r0, c[x].A);                else ans = n - query(r1, i) + query(r1, c[x].A);            }            printf("%d\n",ans);        }    }    return 0;}