HDU1540 Tunnel Warfare(线段树区间合并)

HDU1540  Tunnel Warfare(线段树区间合并)

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

题意：有n个村庄排成一列，一开始村庄都与相邻村庄通过地下通道相连，D代表 把这个村庄与相邻的村庄隔断，R代表修复上一个被隔断的村庄，即与相邻村庄重新相连，Q代表查询这个村庄能与多少个村庄相连，包括自己

思路：线段树区间合并题，ls表示从左边界开始的最长连续区间，rs表示从右边界开始的最长连续区间，maxs代表该区间内的最长连续区间，建树的时候ls,rs,maxs初始化为r-l+1，更新时为单点更新，pushdown时，先让 ls为左子树的ls，rs为右子树的rs，然后如果左子树的ls等于左子树的区间长，则父节点的ls能与右子树的ls相连，rs同理，maxs则为左子树的maxs 右子树的maxs 左子树rs加右子树ls的最大值，查询时，看代码注释

#include<stdio.h>
#include<algorithm>
using namespace std;
const int MAXN=50005;
struct NODE{
	int l,r;
	int ls,rs,maxs;
}segTree[MAXN<<2];
void build(int num,int l,int r)
{
	segTree[num].l=l;
	segTree[num].r=r;
	segTree[num].ls=segTree[num].rs=segTree[num].maxs=r-l+1;
	if(l==r)  return;
	int mid=(l+r)>>1;
	build(num<<1,l,mid);
	build(num<<1|1,mid+1,r);
}
void update(int num,int t,int val)
{
	if(segTree[num].l==segTree[num].r)
	{
		if(val==1) segTree[num].ls=segTree[num].rs=segTree[num].maxs=1;
		else  segTree[num].ls=segTree[num].rs=segTree[num].maxs=0;
		return;
	}
	int mid=(segTree[num].l+segTree[num].r)>>1;
	if(t<=mid) update(num<<1,t,val);
	else update((num<<1)|1,t,val);
	
	segTree[num].ls=segTree[num<<1].ls;
	segTree[num].rs=segTree[(num<<1)|1].rs;
	segTree[num].maxs=max(segTree[num<<1].maxs,segTree[(num<<1)|1].maxs);
	segTree[num].maxs=max(segTree[num].maxs,segTree[num<<1].rs+segTree[(num<<1)|1].ls);
	
	if(segTree[num<<1].ls==segTree[num<<1].r-segTree[num<<1].l+1)
	   segTree[num].ls+=segTree[(num<<1)|1].ls;
	if(segTree[num<<1|1].rs==segTree[num<<1|1].r-segTree[(num<<1)|1].l+1)
	   segTree[num].rs+=segTree[num<<1].rs;      
}
int query(int num,int t)
{
          //多了个剪枝，如果一个区间的最大连续区间为0，或等于区间长度，则返回
         if(segTree[num].l==segTree[num].r||segTree[num].maxs==0||segTree[num].maxs==segTree[num].r-segTree[num].l+1)
	  return segTree[num].maxs;
	int mid=(segTree[num].l+segTree[num].r)>>1;
	
	if(t<=mid)
	{
                   //t在左子树的rs里，能右子树的ls相连
              if(t>=segTree[num<<1].r-segTree[num<<1].rs+1)
		 return query(num<<1,t)+query(num<<1|1,mid+1);
		else return query(num<<1,t);   
	}
	else
	{
                   //同理
              if(t<=segTree[num<<1|1].l+segTree[num<<1|1].ls-1)
		 return query(num<<1|1,t)+query(num<<1,mid);
		else 
		   return query(num<<1|1,t);   
	}    
}
int stack[MAXN];
int main(void)
{
	int n,m;
	int t;
	char op[2];
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		build(1,1,n);
		int top=0;
		while(m--)
		{
			scanf("%s",op);
			if(op[0]=='D')
			{
				scanf("%d",&t);
				stack[top++]=t;
				update(1,t,0);
			}
			else if(op[0]=='Q')
			{
			    scanf("%d",&t);
			    printf("%d\n",query(1,t));
			}
			else{
				if(t>0)
				{
					t=stack[--top];
					update(1,t,1);
				}
			}
		}
	}
	return 0;
}