Luogu P1503 鬼子进村

分块易想难调。
大BUG:
. 块数可能为\(blockSize + 1\)
. 边界为\(\max\{n, block[x] * blockSize\}\)
. 有时块内不行了，就不能去块外，要灵活应对，因题而定
话说这是道平衡树模板题来着

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long

#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 50007;

int n;

int vis[N];
int sta[N], top;
int block[N], blockSize;
int ans[N], peace[N];
inline void Destroy(int x){
	vis[x] = true;
	--ans[block[x]];
}
inline void Rescue(int x){
	vis[x] = false;
	++ans[block[x]];
}
inline int Query(int x){
	if(vis[x] == true) return 0;
	int sum = 0;
	int flagLeft = 0, flagRight = 0;
	int maxx = Min(block[x] * blockSize, n);
	R(i,x + 1,maxx){
		if(vis[i] == true){
			flagRight = 1;
			break;
		}
		++sum;
	}
	nR(i,x - 1, (block[x] - 1) * blockSize + 1){
		if(vis[i] == true){
			flagLeft = 1;
			break;
		}
		++sum;
	}
	if(flagLeft == 1 && flagRight == 1) return sum + 1;
	
	int flag = 0;
	if(flagRight == 0){
		R(i,block[x] + 1, block[n]){
			if(flag) break;
			if(ans[i] == peace[i]){
				sum += ans[i];
				continue;
			}
			flag = 1;
			int maxx = Min(i * blockSize, n);
			R(j,(i - 1) * blockSize + 1, maxx){
				if(vis[j] == true) break;
				++sum;
			}
		}		
	}

	if(flagLeft == 0){
		flag = 0;
		nR(i,block[x] - 1, 1){
			if(flag) break;
			if(ans[i] == peace[i]){
				
				sum += ans[i];
				continue;
			}
			flag = 1;
			nR(j,i * blockSize, (i - 1) * blockSize + 1){
				if(vis[j] == true){
					break;
				}
				++sum;
			}
		}		
	}

	return sum + 1;
}

int main(){
//FileOpen();
	int m;
	io >> n >> m;
	blockSize = sqrt(n) + 1;
	R(i,1,n){
		block[i] = (i - 1) / blockSize + 1;
		++ans[block[i]];
	}
	R(i,1,blockSize) peace[i] = ans[i];
	while(m--){
		char ch = getchar();
		while(ch != 'D' && ch != 'R' && ch != 'Q') ch = getchar();
		if(ch == 'D'){
			int x;
			io >> x;
			Destroy(x);
			sta[++top] = x;
		}
		else if (ch == 'R'){
			if(!top) continue;
			Rescue(sta[top]);
			--top;
		}
		else{
			int x;
			io >> x;
			printf("%d\n", Query(x));
		}
	}
	
	return 0;
}
/*
$1 belong to 1
$2 belong to 1
$3 belong to 2
$4 belong to 2
$5 belong to 3
$6 belong to 3
$7 belong to 4
*/