luogu P1503 鬼子进村

嘟嘟嘟

线段树好题。

其实挺水的，想暴力怎么做：每一次从这个点开始向两边扩，直到遇到第一个摧毁的房屋。

那么把暴力改成倍增，然后线段树查询区间和是否为０。时间复杂度O(nlog²n)。

题解好像有线段树的O(nlogn)的做法，但是特别麻烦，也没怎么看懂。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 5e4 + ;
 inline ll read()
 {
   ll ans = ;
   char ch = getchar(), last = ' ';
   while(!isdigit(ch)) {last = ch; ch = getchar();}
   while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
   if(last == '-') ans = -ans;
   return ans;
 }
 inline void write(ll x)
 {
   if(x < ) x = -x, putchar('-');
   if(x >= ) write(x / );
   putchar(x %  + '');
 }

 int n, m;
 char c[];

 int l[maxn << ], r[maxn << ], sum[maxn << ];
 void build(int L, int R, int now)
 {
   l[now] = L; r[now] = R;
   if(L == R) return;
   int mid = (L + R) >> ;
   build(L, mid, now << );
   build(mid + , R, now <<  | );
 }
 void update(int now, int id, int flg)
 {
   if(l[now] == r[now]) {sum[now] += flg; return;}
   int mid = (l[now] + r[now]) >> ;
   if(id <= mid) update(now << , id, flg);
   else update(now <<  | , id, flg);
   sum[now] = sum[now << ] + sum[now <<  | ];
 }
 int query(int L, int R, int now)
 {
   if(l[now] == L && r[now] == R) return sum[now];
   int mid = (l[now] + r[now]) >> ;
   if(R <= mid) return query(L, R, now << );
   else if(L > mid) return query(L, R, now <<  | );
   else return query(L, mid, now << ) + query(mid + , R, now <<  | );
 }

 int solve(int x)
 {
   if(query(x, x, )) return ;
   int ret = ;
   for(int t = x, i = ; i >= ; --i)
     {
       int j = t + ( << i) - ;
       if(j > n) continue;
       if(!query(t, j, )) ret += ( << i), t += ( << i);
     }
   for(int t = x, i = ; i >= ; --i)
     {
       int j = t - ( << i) + ;
       if(j < ) continue;
       if(!query(j, t, )) ret += ( << i), t -= ( << i);
     }
   return ret - ;
 }

 int st[maxn], top = ;

 int main()
 {
   n = read(); m = read();
   build(, n, );
   for(int i = ; i <= m; ++i)
     {
       scanf("%s", c);
       if(c[] == 'D')
     {
       st[++top] = read();
       update(, st[top], );
     }
       else if(c[] == 'R') update(, st[top--], -);
       else
     {
       int x = read();
       write(solve(x)), enter;
     }
     }
   return ;
 }