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  • bzoj1935: [Shoi2007]Tree 园丁的烦恼lowbit 离散化

    链接

    bzoj
    最好不要去luogu,数据太水

    思路

    一个询问转化成四个矩阵,求起点((0,0)到(x,y))的矩阵
    离线处理,离散化掉y,x不用离散。
    一行一行的求,每次处理完一行之后下一行的贡献直接叠加到当前。
    用lowbit统计

    错误

    离散化小心点,是y-1不是y

    代码

    #include <bits/stdc++.h>
    using namespace std;
    const int N = 1e6 + 7, maxn = 5e5 + 1;
    int read() {
    	int x = 0, f = 1; char s = getchar();
    	for (; s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
    	for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
    	return x * f;
    }
    
    int n, m, ans[N], x[N], y[N], X[N], Y[N];
    pair<int, int> tree[N];
    struct ask {
    	int id, q, x, y;
    	ask(int a = 0, int b = 0, int c = 0, int d = 0) {
    		id = a, q = b, x = c, y = d;
    		if (!x || !y) x = 0, y = 0;
    	}
    	bool operator < (const ask &b) const {
    		return x < b.x;
    	}
    } Q[N<<2];
    
    namespace BIT {
    	int sum[N];
    	inline int lowbit(int x) {
    		return x & -x;
    	}
    	void add(int x) {
    		for (int i = x; i <= maxn; i += lowbit(i))
    			sum[i]++;
    	}
    	int query(int x) {
    		int ans = 0;
    		for (int i = x; i >= 1; i -= lowbit(i))
    			ans += sum[i];
    		return ans;
    	}
    }
    int lsh[N<<3],len;
    
    int main() {
    	n = read(), m = read();
    	for (int i = 1; i <= n; ++i) {
    		tree[i].first = read() + 1, tree[i].second = read() + 1;
    		lsh[++len] = tree[i].second;
    	}
    	int nd = 0;
    	for (int i = 1, js = 0; i <= m; ++i) {
    		int x = read() + 1, y = read() + 1, X = read() + 1, Y = read() + 1;
    		lsh[++len] = Y, lsh[++len] = y - 1;
    		if (X and Y)
    		Q[++nd] = ask(i, 1, X, Y);
    		if (X and (y - 1))
    		Q[++nd] = ask(i, -1, X, y - 1);
    		if ((x - 1) and Y)
    		Q[++nd] = ask(i, -1, x - 1, Y);
    		if ((x - 1) and (y - 1))
    		Q[++nd] = ask(i, 1, x - 1, y - 1);
    	}
    
    	sort(lsh + 1, lsh + 1 + len);
    	len = unique(lsh + 1, lsh + 1 + len) - lsh - 1;
    	for (int i = 1; i <= n; ++i)
    		tree[i].second = lower_bound(lsh + 1, lsh + 1 + len, tree[i].second) - lsh;
    	for (int i = 1; i <= nd; ++i) {
    		Q[i].y = lower_bound(lsh + 1, lsh + 1 + len, Q[i].y) - lsh;
    	}
    	
    	sort(tree + 1, tree + 1 + n);
    	sort(Q + 1, Q + 1 + nd);
    
    	int js = 1;
    	for (int i = 1; i <= nd; ++i) {
    		while (tree[js].first <= Q[i].x && js <= n)
    			BIT::add(tree[js].second),++js;
    		ans[Q[i].id] += Q[i].q * BIT::query(Q[i].y);
    	}
    
    	for (int i = 1; i <= m; ++i)
    		printf("%d
    ", ans[i]);
    
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dsrdsr/p/10978001.html
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