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。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 5e4 + 5; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar('-'); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n, m; 38 char c[2]; 39 40 int l[maxn << 2], r[maxn << 2], sum[maxn << 2]; 41 void build(int L, int R, int now) 42 { 43 l[now] = L; r[now] = R; 44 if(L == R) return; 45 int mid = (L + R) >> 1; 46 build(L, mid, now << 1); 47 build(mid + 1, R, now << 1 | 1); 48 } 49 void update(int now, int id, int flg) 50 { 51 if(l[now] == r[now]) {sum[now] += flg; return;} 52 int mid = (l[now] + r[now]) >> 1; 53 if(id <= mid) update(now << 1, id, flg); 54 else update(now << 1 | 1, id, flg); 55 sum[now] = sum[now << 1] + sum[now << 1 | 1]; 56 } 57 int query(int L, int R, int now) 58 { 59 if(l[now] == L && r[now] == R) return sum[now]; 60 int mid = (l[now] + r[now]) >> 1; 61 if(R <= mid) return query(L, R, now << 1); 62 else if(L > mid) return query(L, R, now << 1 | 1); 63 else return query(L, mid, now << 1) + query(mid + 1, R, now << 1 | 1); 64 } 65 66 int solve(int x) 67 { 68 if(query(x, x, 1)) return 0; 69 int ret = 0; 70 for(int t = x, i = 20; i >= 0; --i) 71 { 72 int j = t + (1 << i) - 1; 73 if(j > n) continue; 74 if(!query(t, j, 1)) ret += (1 << i), t += (1 << i); 75 } 76 for(int t = x, i = 20; i >= 0; --i) 77 { 78 int j = t - (1 << i) + 1; 79 if(j < 1) continue; 80 if(!query(j, t, 1)) ret += (1 << i), t -= (1 << i); 81 } 82 return ret - 1; 83 } 84 85 int st[maxn], top = 0; 86 87 int main() 88 { 89 n = read(); m = read(); 90 build(1, n, 1); 91 for(int i = 1; i <= m; ++i) 92 { 93 scanf("%s", c); 94 if(c[0] == 'D') 95 { 96 st[++top] = read(); 97 update(1, st[top], 1); 98 } 99 else if(c[0] == 'R') update(1, st[top--], -1); 100 else 101 { 102 int x = read(); 103 write(solve(x)), enter; 104 } 105 } 106 return 0; 107 }