BZOJ2957: 楼房重建(分块)

题意

题目链接

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
*/
posted @ 2019-02-07 15:50  自为风月马前卒  阅读(193)  评论(0编辑  收藏

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