Codeforces 524E Rooks and Rectangles 线段树

区域安全的check方法就是， 每行都有哨兵或者每列都有哨兵，然后我们用y建线段树， 维护在每个y上的哨兵的x的最值就好啦。

#include
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair
#define PLI pair
#define PII pair
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, m, k, q;
bool ans[N];

PII p[N];
struct Qus {
    PII a, b;
    bool operator < (const Qus& rhs) const {
        return b < rhs.b;
    }
    int id;
} qus[N];

int a[N << 2];
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
void update(int p, int val, int l, int r, int rt) {
    if(l == r) {
        a[rt] = max(a[rt], val);
        return;
    }
    int mid = l + r >> 1;
    if(p <= mid) update(p, val, lson);
    else update(p, val, rson);
    a[rt] = min(a[rt << 1], a[rt << 1 | 1]);
}
int query(int L, int R, int l, int r, int rt) {
    if(l >= L && r <= R) return a[rt];
    int mid = l + r >> 1;
    if(R <= mid) return query(L, R, lson);
    else if(L > mid) return query(L, R, rson);
    else return min(query(L, R, lson), query(L, R, rson));
}

void solve() {
    sort(p + 1, p + 1 + k);
    sort(qus + 1, qus + 1 + q);
    memset(a, 0, sizeof(a));
    for(int i = 1, j = 1; i <= q; i++) {
        while(j <= k && p[j] <= qus[i].b) update(p[j].se, p[j].fi, 1, m, 1), j++;
        int mn = query(qus[i].a.se, qus[i].b.se, 1, m, 1);
        if(mn >= qus[i].a.fi) ans[qus[i].id] = true;
    }
}

int main() {
    scanf("%d%d%d%d", &n, &m, &k, &q);
    for(int i = 1; i <= k; i++) scanf("%d%d", &p[i].fi, &p[i].se);
    for(int i = 1; i <= q; i++) {
        scanf("%d%d", &qus[i].a.fi, &qus[i].a.se);
        scanf("%d%d", &qus[i].b.fi, &qus[i].b.se);
        qus[i].id = i;
    }
    solve();
    swap(n, m);
    for(int i = 1; i <= k; i++) swap(p[i].fi, p[i].se);
    for(int i = 1; i <= q; i++) {
        swap(qus[i].a.fi, qus[i].a.se);
        swap(qus[i].b.fi, qus[i].b.se);
    }
    solve();
    for(int i = 1; i <= q; i++) printf("%s\n", ans[i] ? "YES" : "NO");
    return 0;
}

/*
*/

 

转载于:https://www.cnblogs.com/CJLHY/p/10437045.html