CF 527C Glass Carving

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstring>
#include <set>
using namespace std;
const int MAXN = 400005;
int init() {
    int rv = 0, fh = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fh = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        rv = (rv<<1) + (rv<<3) + c - '0';
        c = getchar();
    }
    return fh * rv;
}
int w, h, n, maw, mah, cnt1[MAXN], cnt2[MAXN];
set <int> s1, s2;
set <int> :: iterator u, v;
int main() {
    freopen("in.txt", "r", stdin);
    w = init(); h = init(); n = init();
    mah = h; maw = w;
    cnt1[mah]++; cnt2[maw]++;
    s1.insert(h);s2.insert(0);s2.insert(w);s1.insert(0); //要提前把这两个点加上
    for(int i = 1 ; i <= n ; i++) {
        char c;
        scanf(" %c ", &c);
        int t = init();
        if(c == 'H'){
            s1.insert(t);
            u = v =s1.find(t);
            u--; v++;
            cnt1[*v - *u]--;
            cnt1[*v - t]++; cnt1[t - *u]++;
            while(!cnt1[mah]) mah--;   //此处因为mah是不增的，所以这个操作的总复杂度是O(h)
        }else {
            s2.insert(t);
            u = v =s2.find(t);
            u--; v++;
            cnt2[*v - *u]--;
            cnt2[*v - t]++; cnt2[t - *u]++;
            while(!cnt2[maw]) maw--;
        }
        printf("%lld\n", (long long)maw * mah);
    }
    fclose(stdin);
    return 0;
}