hdu 1540 Tunnel Warfare 一个关于线段的故事~~~线段树是我无法言明的伤~

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4722    Accepted Submission(s): 1809

Problem Description

During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village was directly connected with two neighboring ones.

Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately!

 

Input

The first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event.

There are three different events described in different format shown below:

D x: The x-th village was destroyed.

Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.

R: The village destroyed last was rebuilt.

 

Output

Output the answer to each of the Army commanders’ request in order on a separate line.

 

Sample Input

   
   
   
   

    
    
    
    7 9
D 3
D 6
D 5
Q 4
Q 5
R
Q 4
R
Q 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
0
2
4
   
   
   
   

 哎，这题的思路很巧妙。。

要记录单元被损坏的顺序，用一个栈就好了，毁坏就入栈，修复就出栈

说说思路，最难的是查询一个点附近有那些的连接着的区间

这需要在线段树记录三个信息，tlen，llen，rlen，这个记录和 poj 3667 Hotel 记录的意义是相同的 ， tlen表示该节点内最长的可用区间的长度，llen表示最左端数起的区间长度，rlen表示从最右端数起的区间长度

对于一个点，看它是在当前区间的左半还是右半

在左半的话，看看是不是在右端的连续区间内，是的话，还要加上右半区间的左端连续区间。否则的话，只要计算它在左半区间的连接情况即可

在右半的话同理，看看是不是在左端的连续区间内，是的话，还要加上左半区间的右端连续区间。否则的话，只要计算它在右半区间的连接情况即可

 

所以需要时刻维护好每个节点的tlen，llen，rlen，在updata函数中，和 poj 3667 Hotel 的维护是一样的

看代码：

#include <stdio.h>
#define MAX 150100

struct Seg{
	int l,r;
	int ls,rs,ms ;	//分别是记录从左端点起连续的区间数目， 从右端点起连续的区间数目，区间内最大的连续区间 数目 
}st[MAX];

int s[MAX/3] , top = 0;		//模拟栈 

void creat(int l , int r ,int pos)
{
	st[pos].l = l ,st[pos].r = r ;
	st[pos].ls = st[pos].rs = st[pos].ms = r-l+1 ;  
	if(l == r)
	{
		return ;
	}
	int mid = (l+r)>>1 ;
	creat(l,mid,pos<<1) ;
	creat(mid+1,r,pos<<1|1) ;
}
int max(int a , int b)
{
	return a>b?a:b ;
}
void insert(int t , int flag , int pos)
{
	if(st[pos].l == st[pos].r)
	{
		if(flag == 1)
		{
			st[pos].ls = st[pos].rs = st[pos].ms = 1 ;	//修复 
		}
		else
		{
			st[pos].ls = st[pos].rs = st[pos].ms = 0 ;	//毁坏 
		}
		return ;
	}
	int mid = (st[pos].l+st[pos].r)>>1 ;
	if(t<=mid)	insert(t,flag,pos<<1) ;
	else	insert(t,flag,pos<<1|1) ;
	st[pos].ls = st[pos<<1].ls;
	st[pos].rs = st[pos<<1|1].rs ;
	if(st[pos<<1].ls == st[pos<<1].r-st[pos<<1].l+1)
	{
		st[pos].ls += st[pos<<1|1].ls ;
	}
	if(st[pos<<1|1].rs == st[pos<<1|1].r-st[pos<<1|1].l+1)
	{
		st[pos].rs += st[pos<<1].rs ;
	}
	//父亲区间内的最大区间必定是，左子树最大区间，右子树最大区间，左右子树合并的中间区间，三者中最大的区间值
	st[pos].ms = max(max(st[pos<<1].ms,st[pos<<1|1].ms),st[pos<<1].rs+st[pos<<1|1].ls) ;
}
int query(int t , int pos)
{
	if(st[pos].l == st[pos].r || st[pos].ms == 0 || st[pos].ms == st[pos].r-st[pos].l+1)
	{
		return st[pos].ms ;
	}
	int mid = (st[pos].l+st[pos].r)>>1 ;
	if(t<=mid)
	{
		//下面的请注意 
		if(t >= st[pos<<1].r-st[pos<<1].rs+1)//因为t<=mid，看左子树，st[POS<<1].r-st[POS<<1].rs+1代表左子树右边连续区间的左边界值，如果t在左子树的右区间内，则要看右子树的左区间有多长并返回 
		{
			return query(t,pos<<1)+query(mid+1,pos<<1|1) ;
		}
		else
		{
			return query(t,pos<<1) ;
		}
	}
	else
	{
		if(t <= st[pos<<1|1].l+st[pos<<1|1].ls-1)
		{
			return query(mid,pos<<1)+query(t,pos<<1|1) ;
		}
		else
		{
			return query(t,pos<<1|1) ;
		}
	}
}
int main()
{
	int n , m;
	while(~scanf("%d%d",&n,&m))
	{
		char flag[3];
		int t;
		creat(1,n,1) ;
		top = 0 ;
		for(int i = 0 ;i < m ; ++i)
		{
			scanf("%s",flag) ;
			if(flag[0] == 'D')
			{
				scanf("%d",&t) ;
				insert(t,0,1) ;
				s[top++] = t ; 
			}
			else if(flag[0] == 'Q')
			{
				scanf("%d",&t) ;
				printf("%d\n",query(t,1));
			}
			else
			{
				if(top>=0)
				{
					insert(s[--top],1,1) ;
				}
			}
		}
		
	}
	return 0 ;
}

与君共勉