bzoj2002 弹飞绵羊2

分块做法。

因为写的是LCT所以这个思路只是大概了解了一下。

十分脚妙。（23333/斜眼笑）

蒯来的代码，不谢~orz

#include 
#include
#include
#include
#include

using namespace std;

int n,m;
int bl[200011], f[200011], to[200011];
int k[200011];

void init(){
	scanf("%d", &n);
	int sq, r; 
	//??? 
	sq = int(sqrt(n));
	int o = sq, t = 0;
	for(int i = 0; i < n; i++){
		if(o == sq){
			bl[i] = ++t;
			o = 1;
		}
		else bl[i] = t, o++;
		scanf("%d", &f[i]);
		f[i] += i;
	}
	for(int i = n-1; i >= 0; i--){
		r = f[i];
		if(bl[r] != bl[i] || r >= n) to[i] = r, k[i] = 1;
		else{
			to[i] = to[r]; 
			k[i] = k[r] + 1;
		}
	}
}
int ask(int x){
	int he = 0;
	while(x < n){
		  he += k[x]; x = to[x];
	}
	return he;
}
void change(int x, int y){
	int j = bl[x], r;
	f[x] = y; r = y;
	if (bl[r] == bl[x]) to[x] = to[r], k[x] = k[r] + 1;
	else to[x] = y, k[x] = 1;
	for(int i = x-1; i >= 0; i--){
		if(bl[i] != j) break;
		r = f[i];
		if(bl[r] == bl[i] && r < n) to[i] = to[r], k[i] = k[r] + 1;
		else to[i] = r, k[i] = 1;
	}
}
		  
int main(){
	init();
	int j, aa, bb, cc;
	scanf("%d", &m);
	while(m--){
		scanf("%d", &aa);
		if(aa == 1){
			scanf("%d", &bb);
			printf("%d\n", ask(bb));
		}
		else{
			scanf("%d%d", &bb, &cc);
			cc += bb; change(bb, cc);
		} 
	}
	return 0;
}
/*
4 
1 2 1 1 
3
1 1
2 1 1
1 1
*/
/*
2
3
*/