bzoj2716: \[Violet 3\]天使玩偶 cdq分治 曼哈顿距离

bzoj2716: [Violet 3]天使玩偶

Description

Input

Output

HINT

分析

关键在于每个点的答案可以分成四个方向考虑。
所以可以旋转四次坐标系，然后把绝对值去掉。
剩下变成询问一个点下方最大值。
用主席树/cdq分治都可以
似乎确凿是卡了空间的，所以主席树没救了。
老老实实cdq

代码

#include
#include
#include
const int N = 1e6 + 10;
int ri() {
    char c = getchar(); int x = 0, f = 1; for(;c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
    for(;c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) - '0' + c; return x * f;
}
struct D {int x, y, id;}Q[N], q[N], p[N]; int n, m, id, mx, t[N], A[N];
void TA(int x, int a) {for(;x <= mx; x += x&-x) t[x] = ~a ? std::max(t[x], a) : 0;}
int TQ(int x) {int r = -1e9; for(;x; x -= x&-x) if(t[x]) r = std::max(r, t[x]); return r;}
void Solve(int L, int R) {
    if(L == R) return ; int m = L + R >> 1;
    Solve(L, m); Solve(m + 1, R);
    int t1 = L, t2 = m + 1;
    for(int i = L;i <= R; ++i) {
        if(t1 <= m && q[t1].x <= q[t2].x || t2 > R) {
            if(!q[t1].id) TA(q[t1].y, q[t1].x + q[t1].y);
            p[i] = q[t1++];
        }
        else {
            if(q[t2].id) A[q[t2].id] = std::min(A[q[t2].id], q[t2].x + q[t2].y - TQ(q[t2].y));
            p[i] = q[t2++];
        }
    }
    for(int i = L;i <= m; ++i) if(!q[i].id) TA(q[i].y, -1);
    for(int i = L; i <= R; ++i) q[i] = p[i];
}
int Tra(int x, int p) {return ~p ? x : mx + 1 - x;}
void Work(int a, int b) {
    for(int i = 1;i <= m; ++i) q[i].x = Tra(Q[i].x, a), q[i].y = Tra(Q[i].y, b), q[i].id = Q[i].id;
    Solve(1, m);
}
int main() {
    std::memset(A, 0x3f, sizeof(A));
    n = ri(); m = ri() + n;
    for(int i = 1;i <= m; ++i) {
        int op = i <= n ? 1 : ri(); Q[i].x = ri() + 1; Q[i].y = ri() + 1;
        mx = std::max(mx, std::max(Q[i].x, Q[i].y));
        Q[i].id = op == 2 ? ++id : 0;
    }
    Work(1, 1); Work(1, -1); Work(-1, -1); Work(-1, 1);
    for(int i = 1;i <= id; ++i) printf("%d\n", A[i]);
    return 0;
}