bzoj2002 [Hnoi2010]Bounce 弹飞绵羊【分块】

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=2002

这一题除了LCT解法,还有一种更巧妙,代码量更少的解法,就是分块。先想,如果仅仅记录每个节点需要几步可以弹飞,就可以做到O(1)查询O(n)修改;如果仅仅记录每个节点弹力洗漱,就可以做到O(n)查询O(1)修改。这会不会给人一种随机访问数组与链表的感觉呢?如果把n个弹簧分成√n块,记录块里每个弹簧需要几步才能跳出这一个块,并且记录跳出这个块后落到了哪,这样子查询以及修改复杂度都是O(√n)了。

#include <cstdio>
#include <cmath>

const int maxn = 200005;

int n, m, a[maxn], t1, t2, t3, siz, zuihou, kaishi, to[maxn], stp[maxn];

inline int qry(int pos) {
	int rt = 0;
	while (~pos) {
		rt += stp[pos];
		pos = to[pos];
	}
	return rt;
}
inline void upd(int pos, int data) {
	kaishi = pos / siz * siz;
	zuihou = (pos / siz + 1) * siz - 1;
	a[pos] = data;
	for (int i = pos; i >= kaishi; --i) {
		if (i + a[i] >= n) {
			to[i] = -1;
			stp[i] = 1;
		}
		else if (i + a[i] > zuihou) {
			to[i] = i + a[i];
			stp[i] = 1;
		}
		else {
			to[i] = to[i + a[i]];
			stp[i] = stp[i + a[i]] + 1;
		}
	}
}

int main(void) {
	//freopen("in.txt", "r", stdin);
	scanf("%d", &n);
	siz = (int)sqrt((float)n + 0.5f);
	for (int i = 0; i < n; ++i) {
		scanf("%d", a + i);
	}
	for (int i = n - 1; ~i; --i) {
		zuihou = (i / siz + 1) * siz - 1;
		if (i + a[i] >= n) {
			to[i] = -1;
			stp[i] = 1;
		}
		else if (i + a[i] > zuihou) {
			to[i] = i + a[i];
			stp[i] = 1;
		}
		else {
			to[i] = to[i + a[i]];
			stp[i] = stp[i + a[i]] + 1;
		}
	}
	scanf("%d", &m);
	while (m--) {
		scanf("%d%d", &t1, &t2);
		if (t1 == 1) {
			printf("%d\n", qry(t2));
		}
		else {
			scanf("%d", &t3);
			upd(t2, t3);
		}
	}
	return 0;
}

  

posted @ 2016-11-24 19:10  ciao_sora  阅读(1120)  评论(0编辑  收藏