1598E

 通过数组dp[0/1][i][j]表示以（i, j）结尾的楼梯数量，1表示横着的，0表示竖着的

初始值把所有dp[0/1][i][j]赋值为1，但这样会导致所有长度为1的楼梯被统计了两次，故用tot记录长度为1的楼梯

注意考虑line 68 70的情况(边界条件不要忘记重新赋值1)

注意line 77 82 88 89，向后更新时如果后续某个点已经被锁定（不能作为阶梯），则置0

// Decline is inevitable,
// Romance will last forever.
#include <bits/stdc++.h>
using namespace std;
//#define mp make_pair
#define pii pair<int,int>
#define pb push_back
#define fi first
#define se second
#define ll long long
#define LL long long
#define bug(x) cout << x << '\n'
#define int long long
const int maxn = 1e3 + 10;
const int maxm = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int dx[] = {0, 0, -1, 1}; //{0, 0, 1, 1, 1,-1,-1,-1}
const int dy[] = {1, -1, 0, 0}; //{1,-1, 1, 0,-1, 1, 0,-1}
const int P = 1e9 + 7;
int dp[2][maxn][maxn];  //0表示以竖着结尾 1表示横着结尾
int n, m, q;
int ans, tot;
int a[maxn][maxn];
void init() {ans = 0;tot = n * m;for(int i = 1; i <= n; i++)for(int j = 1; j<= m; j++)for(int k = 0; k < 2; k++){dp[k][i][j] = 1;a[i][j] = 1;}for(int i = 1; i <= n; i++)for(int j =1; j <= m; j++) {if(i > 1) dp[0][i][j] = dp[1][i-1][j] + 1;if(j > 1) dp[1][i][j] = dp[0][i][j-1] + 1;}for(int i = 1; i <= n; i++)for(int j =1; j <= m; j++)for(int k = 0; k < 2;k ++)ans += dp[k][i][j]; //初始的ans}
void solve() {cin >> n >> m >> q;init();while(q--) {int x , y;cin >> x >> y;a[x][y] = 1 - a[x][y];
//        cerr << a[x][y] << endl;if(a[x][y] == 0) {tot--;for(int k = 0; k < 2; k++){ans -= dp[k][x][y];dp[k][x][y] = 0;}for(int k = 0; k < 2; k++)ans += dp[k][x][y];}else {tot++;for(int k = 0; k < 2; k++)ans -= dp[k][x][y];if(x > 1) dp[0][x][y] = dp[1][x-1][y] + 1;else dp[0][x][y] = 1;if(y > 1) dp[1][x][y] = dp[0][x][y-1] + 1;else dp[1][x][y] = 1;for(int k = 0; k < 2; k++)ans += dp[k][x][y];}while(x <= n && y <= m) {if(x  < n) {ans -= dp[0][x+1][y];dp[0][x + 1][y] = (dp[1][x][y] + 1) * a[x+1][y];ans += dp[0][x+1][y];}if(y < m) {ans -= dp[1][x][y + 1];dp[1][x][y + 1] = (dp[0][x][y] + 1) * a[x][y+1];ans += dp[1][x][y + 1];}if(x < n && y < m) {for(int k = 0; k < 2; k++)ans -= dp[k][x+1][y+1];dp[1][x+1][y+1] = (dp[0][x+1][y] + 1) * a[x+1][y+1];dp[0][x+1][y+1] = (dp[1][x][y+1] + 1) * a[x+1][y+1];for(int k = 0; k < 2; k++)ans += dp[k][x + 1][y + 1];}x++; y++;}cout << ans - tot << '\n';}
}
signed main() {ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
//    int T; scanf("%d", &T); while(T--)
//    freopen("1.txt","r",stdin);
//    freopen("2.txt","w",stdout);
//    int T; cin >> T;while(T--)solve();return 0;
}
/**/