Luogu P2468 [SDOI2010]粟粟的书架

一道二合一的题目。两部分思维难度都不太高，但是也都很巧妙。尤其是主席树的\(50\)分，由于本人初学主席树，所以没有见过主席树上二分的套路，就被小小的卡了一下。。

\(n <= 200\) \(and\) \(m <= 200\):前缀和+二分

\(n <= 1\) \(and\) \(m <= 500000\):主席树+二分，即对主席树每个节点维护一个权值和和本数和。关于最后的一个小细节，下面图片有讲解：

#include <bits/stdc++.h>
using namespace std;

int n, m, t;

void subtask_2 () {
	const int N = 210, M = 1010;
	static int rec[N][N], sum[N][N][M], num[N][N][M];
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			cin >> rec[i][j];
			for (int k = rec[i][j]; k >= 1; --k) {
				sum[i][j][k] = rec[i][j];  //[1, 1] -> [i, j] 间页数 >= k 的书页数和				
				num[i][j][k] = 1; 		   //[1, 1] -> [i, j] 间页数 >= k 的书本数和 
			}
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			for (int k = 1; k <= 1000; ++k) {
				sum[i][j][k] += sum[i - 1][j][k] + sum[i][j - 1][k] - sum[i - 1][j - 1][k];
				num[i][j][k] += num[i - 1][j][k] + num[i][j - 1][k] - num[i - 1][j - 1][k];
			}
		}
	}
	
	for (int i = 1; i <= t; ++i) {
		static int a1, b1, a2, b2, h;
		cin >> a1 >> b1 >> a2 >> b2 >> h;
		int l = 1, r = 1000;
		while (l < r) {
			int mid = (l + r + 1) >> 1;
			if (sum[a2][b2][mid] - sum[a2][b1 - 1][mid] - sum[a1 - 1][b2][mid] + sum[a1 - 1][b1 - 1][mid] >= h) {
				l = mid;
			} else {
				r = mid - 1;
			}
		}
		int _num = num[a2][b2][l] - num[a2][b1 - 1][l] - num[a1 - 1][b2][l] + num[a1 - 1][b1 - 1][l];
		int _sum = sum[a2][b2][l] - sum[a2][b1 - 1][l] - sum[a1 - 1][b2][l] + sum[a1 - 1][b1 - 1][l];
		while (_sum - l >= h) _sum -= l, _num -= 1;
		if (_sum < h) {
			puts ("Poor QLW");
		} else {
			cout << _num << endl;
		}	
	}
}
const int N = 500010;

static int tot = 0, rt[N], arr[N];

struct Segment_Tree {
	
	struct Segment_Node {
		int ls, rs, sz, sum;
	}t[N << 5];
	
	int modify (int root, int l, int r, int _val) {
		int p = ++tot, mid = (l + r) >> 1;
		t[p].ls = t[root].ls;
		t[p].rs = t[root].rs;
		t[p].sz = t[root].sz + 1;
		t[p].sum = t[root].sum + _val;
		if (l != r) { 
			if (_val <= mid) {
				t[p].ls = modify (t[root].ls, l, mid, _val);
			} else {
				t[p].rs = modify (t[root].rs, mid + 1, r, _val);
			}
		}
		return p;
	}
	
	int build (int l, int r) {
		int p = ++tot, mid = (l + r) >> 1;
		t[p].sz = t[p].sum = 0;
		if (l != r) {
			t[p].ls = build (l, mid);
			t[p].rs = build (mid + 1, r);
		} else {
			t[p].ls = t[p].rs = 0;
		}
		return p;
	}
	
	int query (int u, int v, int l, int r, int k) {
	    int ans = 0;
	    while (l < r) {
	        int mid = (l + r) >> 1;
	        int lch = t[t[v].rs].sum - t[t[u].rs].sum;
	        if (lch < k) {
				ans += t[t[v].rs].sz - t[t[u].rs].sz, k -= lch; 
				r = mid;
				v = t[v].ls, u = t[u].ls;
	        } else {
				l = mid + 1;
				v = t[v].rs, u = t[u].rs;
	    	}
		}
	    ans += ceil ((1.0 * k) / (1.0 * l));
	 	return ans;
	}
	
//	int query (int u, int v, int l, int r, int sumw) {
//		int mid = (l + r) >> 1;
//		int del = t[t[v].ls].sum - t[t[u].ls].sum, ans = 0;
//		if (l != r) {
//			if (sumw <= del) {
//				ans += query (t[u].ls, t[v].ls, l, mid, sumw);
//			} else {
//				ans += del;
//				ans += query (t[u].rs, t[v].rs, mid + 1, r, sumw - del);
//			}
//		}
//		return ans;
//	}
}tree;
void subtask_1 () {

	rt[0] = tree.build (1, 1000);
	for (int i = 1; i <= m; ++i) {
		cin >> arr[i];
		rt[i] = tree.modify (rt[i - 1], 1, 1000, arr[i]);
	}
	for (int i = 1; i <= t; ++i) {
		static int a1, b1, a2, b2, h;
		cin >> a1 >> b1 >> a2 >> b2 >> h;
		if (tree.t[rt[b2]].sum - tree.t[rt[b1 - 1]].sum < h) {
			puts ("Poor QLW");
		} else {
			cout << tree.query (rt[b1 - 1], rt[b2], 1, 1000, h) << endl;
		}
	}
}

int main () {
	freopen ("2468.in", "r", stdin);
	cin >> n >> m >> t;
	if (n == 1) subtask_1 ();
	if (n != 1) subtask_2 ();
}