[CF846D]Monitor题解

看了一眼这题所用的操作，我觉得二维树状数组珂做，然后发现如果按时间顺序把节点一个个加进去再判会TLE，但发现二分时间明显比刚刚的做法快，于是二分时间+暴力插入该时间之内的点+树状数组维护即可AC

贴个代码：

#include <cstdio>
#include <cstring>
#define ll long long
#define lowbit(x) (x&(-x))

inline ll read(){
    ll x = 0; int zf = 1; char ch = ' ';
    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') zf = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;
}

int C[1005][1005];
int n, m, k, q;
int xi[250005], yi[250005];
ll ti[250005];

void clr(){
    memset(C, 0, sizeof(C));
}

void modify(int i, int j, int val){
    for(int x = i; x <= n; x += lowbit(x))
        for(int y = j; y <= m; y += lowbit(y))
            C[x][y] += val;
}

int getsum(int i, int j){
   int res = 0;
   for(int x = i; x > 0; x -= lowbit(x))
        for(int y = j; y > 0; y -= lowbit(y))
            res += C[x][y];
   return res;
}

inline bool judge(ll num){
    clr();
    for (int i = 1; i <= q; ++i)
        if (ti[i] <= num)
            modify(xi[i], yi[i], 1);
    for (int i = k; i <= n; ++i)
        for (int j = k; j <= m; ++j){
            int tmp1 = getsum(i, j) - getsum(i - k, j) - getsum(i, j - k) + getsum(i - k, j - k);
            if (tmp1 == k * k)
                return true;
        }
    return false;
}

int main(){
    n = read(), m = read(), k = read(), q = read();
    ll _max = 0;
    for (int i = 1; i <= q; ++i){
        xi[i] = read(), yi[i] = read(), ti[i] = read();
        if (ti[i] > _max) _max = ti[i];
    }
    ll l = 0, r = _max + 1, ans = -1;
    while (l <= r){
        ll mid = (l + r) >> 1ll;
        if (judge(mid)){
            r = mid - 1;
            ans = mid;
        }
        else
            l = mid + 1;
    }
    printf("%d", ans);
    return 0;
}