【BZOJ1176】[BOI2007]Mokia 摩基亚

【BZOJ1176】[BOI2007]Mokia 摩基亚

题面

bzoj

洛谷

题解

显然的\(CDQ\)\(/\)树套树题

然而根本不想写树套树，那就用\(CDQ\)吧。。。

考虑到点\((x1,y1)\)和\((x2,y2)\)区域内既有上限又有下限我们不是很好算

于是将这个区域的贡献写成另外一种形式，

记\((x,y)\)与\((0,0)\)之间区域的贡献为\(S_{(x,y)}\)，则上面的贡献可表示为

\(S_{(x2,y2)}-S_{(x2,y1-1)}-S_{(y2,x1-1)}+S_{(x1-1,y1-1)}\)

分别统计四个贡献就很容易辣

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
namespace IO {
    const int BUFSIZE = 1 << 20;
    char ibuf[BUFSIZE], *is = ibuf, *it = ibuf;
    inline char gc() {
        if (is == it) it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin);
		return *is++;
    }
}
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (ch != '-' && (ch > '9' || ch < '0')) ch = IO::gc();
    if (ch == '-') w = -1 , ch = IO::gc();
    while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = IO::gc();
    return w * data;
}
#define MAX_N 200005
#define MAX_W 2000005
struct Node { int x, y, id, w; } t[MAX_N << 2]; int ans[MAX_N];
bool cmp_x(Node a, Node b) { return (a.x == b.x) ? (a.y < b.y) : (a.x < b.x); } 

int N, c[MAX_W];
inline int lb(int x) { return x & -x; }
void add(int x, int v) { while (x <= N) c[x] += v, x += lb(x); }
int sum(int x) { int res = 0; while (x > 0) res += c[x], x -= lb(x); return res; }
void Div(int l, int r) {
    if (l == r) return ;
    int mid = (l + r) >> 1;
    Div(l, mid); Div(mid + 1, r);
    int j = l;
    for (int i = mid + 1; i <= r; i++) {
        if (t[i].id == 0) continue;
        for (; t[j].x <= t[i].x && j <= mid; ++j) if (t[j].id == 0) add(t[j].y, t[j].w);
        ans[t[i].id] += sum(t[i].y) * t[i].w;
    }
    for (int i = l; i < j; i++) if (t[i].id == 0) add(t[i].y, -t[i].w);
    inplace_merge(&t[l], &t[mid + 1], &t[r + 1], cmp_x);
}
int main () {
    gi(); N = gi(); int tot = 0, cnt = 0;
    for (;;) {
        int op = gi(); if (op == 3) break;
        int x = gi(), y = gi();
		if (op == 1) t[++tot] = (Node){x, y, 0, gi()};
		else {
		    int _x = gi(), _y = gi(); ++cnt;
		    t[++tot] = (Node){_x, _y, cnt, 1};
		    t[++tot] = (Node){_x, y - 1, cnt, -1};
		    t[++tot] = (Node){x - 1, _y, cnt, -1};
		    t[++tot] = (Node){x - 1, y - 1, cnt, 1};
	    }
    }
    Div(1, tot);
    for (int i = 1; i <= cnt; i++) printf("%d\n", ans[i]);
    return 0;
}