标签:过程 之间 char include push zoj 接下来 print 并且
题目描述
输入
输出
对于每个bridge命令与excursion命令,输出一行,为题目描述所示。
样例输入
5
4 2 4 5 6
10
excursion 1 1
excursion 1 2
bridge 1 2
excursion 1 2
bridge 3 4
bridge 3 5
excursion 4 5
bridge 1 3
excursion 2 4
excursion 2 5
样例输出
4
题解
明知道可以并查集+树剖却偏要使用LCT,为啥?因为好写啊~
在LCT的Splay Tree上维护一个sum,表示实子树的点权和。
然后模拟操作就行了,修改时直接Splay后修改就行。
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 30010
using namespace std;
int fa[N] , c[2][N] , w[N] , sum[N] , rev[N];
char str[15];
void pushup(int x)
{
	sum[x] = sum[c[0][x]] + sum[c[1][x]] + w[x];
}
void pushdown(int x)
{
	if(rev[x])
	{
		int l = c[0][x] , r = c[1][x];
		swap(c[0][l] , c[1][l]) , swap(c[0][r] , c[1][r]);
		rev[l] ^= 1 , rev[r] ^= 1 , rev[x] = 0;
	}
}
bool isroot(int x)
{
	return c[0][fa[x]] != x && c[1][fa[x]] != x;
}
void update(int x)
{
	if(!isroot(x)) update(fa[x]);
	pushdown(x);
}
void rotate(int x)
{
	int y = fa[x] , z = fa[y] , l = (c[1][y] == x) , r = l ^ 1;
	if(!isroot(y)) c[c[1][z] == y][z] = x;
	fa[x] = z , fa[y] = x , fa[c[r][x]] = y , c[l][y] = c[r][x] , c[r][x] = y;
	pushup(y) , pushup(x);
}
void splay(int x)
{
	update(x);
	while(!isroot(x))
	{
		int y = fa[x] , z = fa[y];
		if(!isroot(y)) rotate((c[0][y] == x) ^ (c[0][z] == y) ? x : y);
		rotate(x);
	}
}
void access(int x)
{
	int t = 0;
	while(x) splay(x) , c[1][x] = t , pushup(x) , t = x , x = fa[x];
}
int find(int x)
{
	access(x) , splay(x);
	while(c[0][x]) pushdown(x) , x = c[0][x];
	return x;
}
void makeroot(int x)
{
	access(x) , splay(x) , swap(c[0][x] , c[1][x]) , rev[x] ^= 1;
}
void link(int x , int y)
{
	makeroot(x) , fa[x] = y;
}
int main()
{
	int n , i , m , x , y;
	scanf("%d" , &n);
	for(i = 1 ; i <= n ; i ++ ) scanf("%d" , &w[i]) , sum[i] = w[i];
	scanf("%d" , &m);
	while(m -- )
	{
		scanf("%s%d%d" , str , &x , &y);
		if(str[0] == ‘b‘)
		{
			if(find(x) == find(y)) puts("no");
			else puts("yes") , link(x , y);
		}
		else if(str[0] == ‘p‘) splay(x) , w[x] = y , pushup(x);
		else
		{
			if(find(x) != find(y)) puts("impossible");
			else makeroot(x) , access(y) , splay(y) , printf("%d\n" , sum[y]);
		}
	}
	return 0;
}
标签:过程 之间 char include push zoj 接下来 print 并且
原文地址:http://www.cnblogs.com/GXZlegend/p/7189990.html