BZOJ2957: 楼房重建(分块)

题意

Sol

自己YY出了一个\(n \sqrt{n} \log n\)的辣鸡做法没想到还能过。。

可以直接对序列分块，我们记第\(i\)个位置的值为\(a[i] = \frac{H_i}{i}\)，那么显然一个位置能被看到当前仅当前面的\(a[i]\)都比他小。可以直接拿个vector维护，每次暴力在vector里二分

#include<bits/stdc++.h>

using namespace std;

const int MAXN = 1e5 + 10;

inline int read() {

	char c = getchar(); int x = 0, f = 1;

	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

	return x * f;

}

int N, M, block, belong[MAXN], ll[MAXN], rr[MAXN], lim;

double mx[MAXN], a[MAXN];

vector<double> v[MAXN];

void rebuild(int k, int p, int val) {

	int l = ll[k], r = rr[k]; a[p] = (double) val / p;

	v[k].clear(); mx[k] = 0;

	for(int i = l; i <= r; i++) mx[k] = max(mx[k], a[i]);

	sort(v[k].begin(), v[k].end());

	double cur = 0;

	for(int i = l; i <= r; i++) {

		if(a[i] > cur) v[k].push_back(a[i]);

		cur = max(cur, a[i]);

	}

}

int calc() {

	int ret = 0; double cur = 0;

	for(int i = 1; i <= lim; i++) {

		ret += (v[i].size() - (upper_bound(v[i].begin(), v[i].end(), cur) - v[i].begin()));

		cur = max(cur, mx[i]);

	}

	return ret;

}

int main() {

	N = read(); M = read(); block = sqrt(N *log2(N));

	for(int i = 1; i <= N; i++) belong[i] = (i - 1) / block + 1, lim = max(lim, belong[i]);

	for(int i = 1; i <= lim; i++) ll[i] = (i - 1) * block + 1, rr[i] = ll[i] + block - 1;

	for(int i = 1; i <= M; i++) {

		int x = read(), y = read();

		rebuild(belong[x], x, y);

		printf("%d\n", calc());

	}

	return 0;

}

/*

3 4

2 4

3 6

1 1000000000

1 1

*/

巴特西

BZOJ2957: 楼房重建(分块)

题意

Sol

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