BZOJ2683 简单题（CDQ分治）

传送门

之前听别人说CDQ分治不难学，今天才知道果真如此。之前一直为自己想不到CDQ的方法二很不爽，今天终于是想出来了一道了，太弱……

cdq分治主要就是把整段区间分成两半，然后用左区间的值去更新右区间的答案，每次把区间折半。对于本题来说时间复杂度T(N)=T(N/2)+O(NlogN)

T(N)=O(Nlog2N)

/**************************************************************
    Problem: 2683
    Language: C++
    Result: Accepted
    Time:6204 ms
    Memory:35184 kb
****************************************************************/

#include <cstdio>
#include <algorithm>
using namespace std;
#define MAXN 200005
struct Node {
    int t, x, y, v, k;
    inline bool operator < (const Node&r) const {
        if(x == r.x && y == r.y) return t < r.t;
        else if(x == r.x) return y < r.y;
        else return x < r.x;
    }
} q[MAXN << 2], nq[MAXN << 2];
int n, cnt, ans;
inline void GET(int &n) {
    n = 0; char c;
    do c = getchar(); while('0' > c || c > '9');
    do n = n * 10 + c - '0', c = getchar(); while('0' <= c && c <= '9');
}
namespace BIT {
    int t[MAXN << 2];
    inline void Add(int x, int v) {
        for(; x <= n; x += x&-x) t[x] += v;
    }
    inline int Query(int x, int s = 0) {
        for(; x; x -= x&-x) s+=t[x];
        return s;
    }
}
void cdq(int l, int r) {
    if(l >= r) return;
    using namespace BIT;
    int mid = (l + r) >> 1, lp = l, rp = mid+1;
    for(int i = l; i <= r; ++ i)
        if(q[i].t <= mid && q[i].k == 1) Add(q[i].y, q[i].v);
        else if(q[i].t > mid && q[i].k == 2) q[i].v += Query(q[i].y);
    for(int i = l; i <= r; ++ i) {
        if(q[i].t <= mid && q[i].k == 1) Add(q[i].y, -q[i].v);
        if(q[i].t <= mid) nq[lp ++] = q[i];
        else nq[rp ++] = q[i];
    }
    for(int i = l; i <= r; ++ i) q[i] = nq[i];
    cdq(l, mid); cdq(mid+1, r);
}
int main() {
    scanf("%d", &n);
    int op, x, y, a, b;
    while(~scanf("%d", &op) && 3 != op) {
        if(1 == op) {
            GET(x); GET(y); GET(a);
            q[++ cnt] = { cnt, x, y, a, 1 };
        }
        else {
            GET(x); GET(y); GET(a); GET(b);
            q[++ cnt] = { cnt, x-1, y-1, 0, 2 };
            q[++ cnt] = { cnt, x-1, b, 0, 2 };
            q[++ cnt] = { cnt, a, y-1, 0, 2 };
            q[++ cnt] = { cnt, a, b, 0, 2 };
        }
    }
    sort(q+1, q+cnt+1);
    cdq(1, cnt);
    for(int i = 1; i <= cnt; ++ i)
        if(q[i].k == 2) {
            ans = q[i].v - q[i+1].v - q[i+2].v + q[i+3].v;
            printf("%d\n", ans); i += 3;
        }
    return 0;
}