SEERC 2018 B. Broken Watch （CDQ分治）

题目链接：http://codeforces.com/gym/101964/problem/B

题意：q 种操作，①在(x,y)处加一个点，②加一个矩阵{(x1,y1),(x2,y2)}，问每次操作后点在矩阵中或矩阵边界上的对数有多少。

题解：裸的CDQ分治，考虑对点和矩阵分别进行CDQ分治，因为x,y <= 1e9，要将其中一个坐标离散化。要注意先加点和先加矩阵的两种情况下，矩阵的差分边界是不同的。

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define mst(a,b) memset((a),(b),sizeof(a))
#define mp(a,b) make_pair(a,b)
#define pi acos(-1)
#define pii pair<int,int>
#define pb push_back
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int MAXN = 1e5 + 10;
const int MAXM = 1e8 + 10;
const ll mod = 1e9 + 7;

struct node {
    int op, x1, y1, x2, y2, id;
    bool operator < (const node &rhs) {
        if(x1 != rhs.x1)
            return x1 < rhs.x1;
        return id < rhs.id;
    }
} a[MAXN], b[MAXN << 2], c[MAXN << 2];

int tot;

void addnode(int op, int x1, int y1, int id) {
    ++tot;
    b[tot].op = op, b[tot].x1 = x1, b[tot].y1 = y1, b[tot].id = id;
}

int y[MAXN << 1], ysz = 0;
ll ans[MAXN];

int lowbit(int x) {
    return x & (-x);
}

ll bit[MAXN << 1];

void add(int y, int val) {
    while(y <= ysz) {
        bit[y] += val;
        y += lowbit(y);
    }
}

ll query(int y) {
    ll sum = 0;
    while(y) {
        sum += bit[y];
        y -= lowbit(y);
    }
    return sum;
}

void cdq(int l, int r) {
    if(l == r)
        return ;
    int mid = (l + r) >> 1;
    cdq(l, mid);
    cdq(mid + 1, r);
    int len = 0;
    for(int i = l; i <= mid; i++)
        if(b[i].id == 0)
            c[++len] = b[i];
    for(int i = mid + 1; i <= r; i++)
        if(b[i].id)
            c[++len] = b[i];
    if(len == 0 || c[1].id || c[len].id == 0)
        return ;
    sort(c + 1, c + 1 + len);
    for(int i = 1; i <= len; i++) {
        if(c[i].id == 0)
            add(c[i].y1, c[i].op);
        else
            ans[c[i].id] += 1ll * c[i].op * query(c[i].y1);
    }
    for(int i = 1; i <= len; i++)
        if(c[i].id == 0)
            add(c[i].y1, -c[i].op);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    mst(ans, 0);
    int q;
    scanf("%d", &q);
    for(int i = 1; i <= q; i++) {
        scanf("%d%d%d", &a[i].op, &a[i].x1, &a[i].y1);
        y[++ysz] = a[i].y1;
        if(a[i].op == 2) {
            scanf("%d%d", &a[i].x2, &a[i].y2);
            y[++ysz] = a[i].y2;
        }
    }
    sort(y + 1, y + 1 + ysz);
    ysz = unique(y + 1, y + 1 + ysz) - y - 1;
    for(int i = 1; i <= q; i++) {
        a[i].y1 = lower_bound(y + 1, y + 1 + ysz, a[i].y1) - y;
        if(a[i].op == 2)
            a[i].y2 = lower_bound(y + 1, y + 1 + ysz, a[i].y2) - y;
    }
    tot = 0;
    for(int i = 1; i <= q; i++) {
        if(a[i].op == 1) {
            addnode(1, a[i].x1, a[i].y1, 0);
        } else {
            addnode(1, a[i].x1 - 1, a[i].y1 - 1, i);
            addnode(-1, a[i].x2, a[i].y1 - 1, i);
            addnode(-1, a[i].x1 - 1, a[i].y2, i);
            addnode(1, a[i].x2, a[i].y2, i);
        }
    }
    cdq(1, tot);
    tot = 0;
    for(int i = 1; i <= q; i++) {
        if(a[i].op == 1) {
            addnode(1, a[i].x1, a[i].y1, i);
        } else {
            addnode(1, a[i].x1, a[i].y1, 0);
            addnode(-1, a[i].x2 + 1, a[i].y1, 0);
            addnode(-1, a[i].x1, a[i].y2 + 1, 0);
            addnode(1, a[i].x2 + 1, a[i].y2 + 1, 0);
        }
    }
    cdq(1, tot);
    for(int i = 1; i <= q; i++) {
        ans[i] += ans[i - 1];
        printf("%lld
", ans[i]);
    }
    return 0;
}