[BZOJ2716][Violet 3]天使玩偶 CDQ分治+樹狀數組

按時間分治，把每個詢問拆成四個方向的查詢，這樣曼哈頓距離可以直接用減法得到
一維時間分治，二維x座標排序，三維y座標樹狀數組

#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 1000000 + 10;
const int INF = 1000000000;

int max_x;
int ans[maxn];

struct BIT {
    int c[maxn];
    int lowbit(int x) {
        return x & (-x);
    }
    void modify(int x, int d) {
        while(x <= max_x) {
            c[x] = max(c[x], d);
            x += lowbit(x);
        }
    }
    int query(int x) {
        int ret = 0;
        while(x > 0) {
            ret = max(ret, c[x]);
            x -= lowbit(x);
        }
        return ret;
    }
    void clear(int x) {
        while(x <= max_x) {
            c[x] = 0;
            x += lowbit(x);
        }
    }
} bit;

struct Node {
    int x, y, k, id;
    bool operator < (const Node& rhs) const {
        if(x != rhs.x) return x < rhs.x;
        return id < rhs.id;
    }
} A[maxn];

#define a A[i]
#define b A[j]

struct CDQ {
    int n;
    Node T[maxn];

    void init(int n) {
        this->n = n;
        sort(A+1, A+n+1);
    }

    void solve(int L, int R) {
        if(L >= R) return;
        int M = (L+R) >> 1;
        int i, j, p = L, q = M+1;
        for(i = L; i <= R; i++) if(a.id <= M) T[p++] = a; else T[q++] = a;
        for(i = L; i <= R; i++) A[i] = T[i];
        solve(L, M);
        i = M+1; j = L;
        for(; i <= R; i++) if(a.k == 2) {
            for(; j <= M && b.x <= a.x; j++) if(b.k == 1)
                bit.modify(b.y, b.x+b.y);
            int t = bit.query(a.y);
            if(t) ans[a.id] = min(ans[a.id], a.x+a.y-t);
        }
        for(i = L; i < j; i++) if(a.k == 1)
            bit.clear(a.y);
        solve(M+1, R);
        merge(A+L, A+M+1, A+M+1, A+R+1, T+L);
        for(i = L; i <= R; i++) A[i] = T[i];
    }

} cdq;

int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; i++) {
        scanf("%d %d", &a.x, &a.y);
        a.x++; a.y++; a.id = i; a.k = 1;
        max_x = max(max_x, max(a.x, a.y));
    }
    for(int i = n+1; i <= n+m; i++) {
        scanf("%d %d %d", &a.k, &a.x, &a.y);
        a.x++; a.y++; a.id = i;
        max_x = max(max_x, max(a.x, a.y));
    }
    max_x++; n += m;
    for(int i = 1; i <= n; i++) ans[i] = INF;

    cdq.init(n); cdq.solve(1, n);

    for(int i = 1; i <= n; i++) a.x = max_x - a.x;
    cdq.init(n); cdq.solve(1, n);

    for(int i = 1; i <= n; i++) a.y = max_x - a.y;
    cdq.init(n); cdq.solve(1, n);

    for(int i = 1; i <= n; i++) a.x = max_x - a.x;
    cdq.init(n); cdq.solve(1, n);

    for(int i = 1; i <= n; i++)
        if(ans[i] != INF) printf("%d\n", ans[i]);
    return 0;
}