Tunnel Warfare

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

这题可以用线段树做，数据区间合并，这里要注意题意，每次R恢复的是上次炸过的地方，不管是不是重复炸过，如D 6 D 3 D 3 D 3 R R，那么6那个点并没有恢复，这里WA了快10次。。。
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 50005
char s[10];
int a[maxn],sum,vis[maxn];
struct node{
	int l,r,llen,rlen,tlen;
}b[4*maxn];

int max(int a,int b){
	return a>b?a:b;
}

void build(int l,int r,int i)
{
	int mid;
	b[i].l=l;b[i].r=r;b[i].llen=b[i].rlen=b[i].tlen=(b[i].r-b[i].l+1);
	if(l==r)return;
	mid=(l+r)/2;
	build(l,mid,i*2);
	build(mid+1,r,i*2+1);
}

void update(int id,int panduan,int i)
{
	int mid;
	if(b[i].l==b[i].r){
		if(panduan==1){
			b[i].llen=b[i].tlen=b[i].rlen=0;
		}
		else{
			b[i].llen=b[i].tlen=b[i].rlen=1;
		}
		return;
	}
	if(b[i*2].r>=id)update(id,panduan,i*2);
	else update(id,panduan,i*2+1);
	b[i].tlen=b[i*2].rlen+b[i*2+1].llen;
	b[i].llen=b[i*2].llen;b[i].rlen=b[i*2+1].rlen;
	if(b[i*2].llen==b[i*2].r-b[i*2].l+1){
		b[i].llen+=b[i*2+1].llen;
	}
	if(b[i*2+1].rlen==b[i*2+1].r-b[i*2+1].l+1){
		b[i].rlen+=b[i*2].rlen;
	}
}

int question(int id,int i)
{
	int mid;
	if(b[i].l==b[i].r || b[i].tlen==(b[i].r-b[i].l+1) || b[i].tlen==0){
		return b[i].tlen;
	}
	if(b[i*2].r>=id){
		if(b[i*2].r-b[i*2].rlen+1<=id){
			return b[i*2].rlen+b[i*2+1].llen;
		}
		else return question(id,i*2);
	}
	else{
		if(b[i*2+1].l+b[i*2+1].llen-1>=id){
			return b[i*2].rlen+b[i*2+1].llen;
		}
		else return question(id,i*2+1);
	}
}

int main()
{
	int n,m,i,j,c,num,flag;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		build(1,n,1);
		num=0;
		memset(vis,0,sizeof(vis));
		for(i=1;i<=m;i++){
			scanf("%s",s);
			if(s[0]=='D'){
				scanf("%d",&c);
					a[++num]=c;
                    update(c,1,1);
			}
			else if(s[0]=='R'){
				if(num>=1)update(a[num--],2,1);
			}
			else if(s[0]=='Q'){
				scanf("%d",&c);
				flag=0;
				printf("%d\n",question(c,1));
			}
		}
	}
	return 0;
}