hdu1540-Tunnel Warfare （线段树区间合并）

题意：n个村庄，有三种操作，D x 破坏位置为x的村庄，R 修复上一次被破坏的村庄，Q x 输出含有x村庄的连续村庄的最大个数。线段树搞之，区间合并。

ls[maxn]为当前节点左面的连续区间，rs[maxn]为当前节点左面的连续区间,ms[maxn]当前节点的最大连续区间。

 #include <set>
 #include <map>
 #include <cmath>
 #include <ctime>
 #include <queue>
 #include <stack>
 #include <cctype>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 typedef unsigned long long ull;
 typedef long long ll;
 const int inf = 0x3f3f3f3f;
 const double eps = 1e-;
 const int maxn = 5e4+;
 int ls[maxn<<],ms[maxn<<],rs[maxn<<];
 stack<int>S;
 void build(int l,int r,int pos)
 {
     ls[pos] = rs[pos] = ms[pos] = r - l + ;
     if (l == r)
         return;
     int mid = (l + r) >> ;
     build(l,mid,pos<<);
     build(mid+,r,pos<<|);
 }
 void update(int l,int r,int pos,int x,int val)
 {
     if (l == r)
     {
         ls[pos] = rs[pos] = ms[pos] = val;
         return;
     }
     int mid = (l + r) >> ;
     if (x <= mid)
         update(l,mid,pos<<,x,val);
     if (x > mid)
         update(mid+,r,pos<<|,x,val);
     ms[pos] = max(rs[pos<<]+ls[pos<<|],max(ms[pos<<],ms[pos<<|]));
     ls[pos] = ls[pos<<];
     rs[pos] = rs[pos<<|];
     if (mid-l+==rs[pos<<])
         ls[pos] += ls[pos<<|];
     if (r-mid==ls[pos<<|])
         rs[pos] += rs[pos<<];
 }
 int query(int l,int r,int pos,int x)
 {
     if (ms[pos] == r - l +  || !ms[pos] || l == r)
     {
         return ms[pos];
     }
     int mid = (l + r) >> ;
     if (x <= mid)
     {
         if (x >= mid - rs[pos<<] + )
             return query(l,mid,pos<<,x) + query(mid+,r,pos<<|,mid+);
         else
             return query(l,mid,pos<<,x);
     }
     else
     {
         if (x <= mid+ls[pos<<|])
             return query(mid+,r,pos<<|,x) + query(l,mid,pos<<,mid);
         else
             return query(mid+,r,pos<<|,x);
     }
 }
 int main(void)
 {
     #ifndef ONLINE_JUDGE
         freopen("in.txt","r",stdin);
     #endif
     int n,m;
     while (~scanf ("%d%d",&n,&m))
     {
         build(,n,);
         char op[];
         for (int i = ; i < m; i++)
         {
             int x;
             scanf ("%s",op);
             if (op[] == 'D')
             {
                 scanf ("%d",&x);
                 update(,n,,x,);
                 S.push(x);
             }
             if (op[] == 'Q')
             {
                 scanf ("%d",&x);
                 printf("%d\n",query(,n,,x));
             }
             if (op[] == 'R')
             {
                 if (!S.empty())
                 {
                     int tmp = S.top();
                     S.pop();
                     update(,n,,tmp,);
                 }
             }
         }
         while (!S.empty())
             S.pop();
     }
     return ;
 }