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
思路：
类似这题：（可以一起做了，）
https://www.cnblogs.com/qieqiemin/p/11385427.html
线段树处理连续区间的经典问题。
我们定义一个节点被轰炸为0，没被轰炸为1.
线段树每一个区间维护以下信息：
1、从左端点开始的最长连续1的个数。
2、从右端点开始的最长连续1的个数。
3、整个区间的最长连续1的个数。
4、是否整个区间是否都为1。
具体的转移和pushup看代码即可。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 50010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
struct node
{
int l,r;
int num;
int isall;
int lm,rm;
}segmeng_tree[maxn<<2];
int n;
int m;
void pushup(int rt)
{
if(segmeng_tree[rt<<1].isall)
{
segmeng_tree[rt].isall=segmeng_tree[rt<<1|1].isall;
segmeng_tree[rt].lm=segmeng_tree[rt<<1].lm+segmeng_tree[rt<<1|1].lm;
if(segmeng_tree[rt<<1|1].isall)
segmeng_tree[rt].rm=segmeng_tree[rt<<1|1].rm+segmeng_tree[rt<<1].rm;
else
segmeng_tree[rt].rm=segmeng_tree[rt<<1|1].rm;
segmeng_tree[rt].num=max(segmeng_tree[rt].lm,segmeng_tree[rt].rm);
segmeng_tree[rt].num=max(segmeng_tree[rt].num,segmeng_tree[rt<<1].rm+segmeng_tree[rt<<1|1].lm);
}else if(segmeng_tree[rt<<1|1].isall)
{
segmeng_tree[rt].isall=segmeng_tree[rt<<1].isall;
segmeng_tree[rt].rm=segmeng_tree[rt<<1|1].rm+segmeng_tree[rt<<1].rm;
if(segmeng_tree[rt<<1].isall)
segmeng_tree[rt].lm=segmeng_tree[rt<<1|1].lm+segmeng_tree[rt<<1].lm;
else
segmeng_tree[rt].lm=segmeng_tree[rt<<1].lm;
segmeng_tree[rt].num=max(segmeng_tree[rt].lm,segmeng_tree[rt].rm);
segmeng_tree[rt].num=max(segmeng_tree[rt].num,segmeng_tree[rt<<1].rm+segmeng_tree[rt<<1|1].lm);
}else
{
segmeng_tree[rt].isall=0;
segmeng_tree[rt].lm=segmeng_tree[rt<<1].lm;
segmeng_tree[rt].rm=segmeng_tree[rt<<1|1].rm;
segmeng_tree[rt].num=max(segmeng_tree[rt].lm,segmeng_tree[rt].rm);
segmeng_tree[rt].num=max(segmeng_tree[rt].num,segmeng_tree[rt<<1].rm+segmeng_tree[rt<<1|1].lm);
}
}
void build(int rt,int l,int r)
{
segmeng_tree[rt].l=l;
segmeng_tree[rt].r=r;
if(l==r)
{
segmeng_tree[rt].num=segmeng_tree[rt].lm=segmeng_tree[rt].rm=segmeng_tree[rt].isall=1;
return;
}
int mid=(l+r)>>1;
build(rt<<1,l,mid);
build(rt<<1|1,mid+1,r);
pushup(rt);
}
void update(int rt,int x,int val)
{
if(segmeng_tree[rt].l==segmeng_tree[rt].r)
{
if(val)
{
segmeng_tree[rt].isall=1;
}else
{
segmeng_tree[rt].isall=0;
}
segmeng_tree[rt].lm=segmeng_tree[rt].rm=segmeng_tree[rt].num=val;
}else
{
int mid=(segmeng_tree[rt].l+segmeng_tree[rt].r)>>1;
if(x<=mid)
{
update(rt<<1,x,val);
}else
{
update(rt<<1|1,x,val);
}
pushup(rt);
}
}
int ask(int rt,int x)
{
if(segmeng_tree[rt].l==segmeng_tree[rt].r||segmeng_tree[rt].isall||segmeng_tree[rt].num==0)
{
return segmeng_tree[rt].num;
}
int mid=(segmeng_tree[rt].l+segmeng_tree[rt].r)>>1;
if(x<=mid)
{
if(segmeng_tree[rt<<1].r-segmeng_tree[rt<<1].rm+1<=x)
{
return ask(rt<<1,x)+ask(rt<<1|1,mid+1);
}else
{
return ask(rt<<1,x);
}
}else
{
if(segmeng_tree[rt<<1|1].l+segmeng_tree[rt<<1|1].lm-1>=x)
{
return ask(rt<<1|1,x)+ask(rt<<1,mid);
}else
{
return ask(rt<<1|1,x);
}
}
}
stack<int> st;
int main()
{
//freopen("D:\code\text\input.txt","r",stdin);
//freopen("D:\code\text\output.txt","w",stdout);
while(~scanf("%d %d",&n,&m))
{
build(1,1,n);
while(!st.empty())
{
st.pop();
}
char s[22];
while(m--)
{
scanf("%s",s);
int x;
if(s[0]=='D')
{
scanf("%d",&x);
update(1,x,0);
st.push(x);
}else if(s[0]=='Q')
{
scanf("%d",&x);
printf("%d
",ask(1,x));
}else
{
if(st.size())
{
x=st.top();
st.pop();
update(1,x,1);
}
}
}
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '
');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}