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  • Luogu2801 教主的魔法 (分块)

    与hzw的分块2类似,放vector排序

    #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("
    
    ----------
    
    ")
    #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;
    #include <vector>
    
    const int N = 1000007;
    
    int n;
    
    int block[N], blockSize;
    int a[N], tag[N];
    vector<int> V[1007];
    
    inline void Reset(int x){
    	V[x].clear();
    	R(i,(x - 1) * blockSize + 1, block[x] * blockSize){
    		V[x].push_back(a[i]);
    	}
    	sort(V[x].begin(), V[x].end());
    }
    inline void Updata(int L, int R, int w){
    	int minn = Min(R, block[L] * blockSize);
    	R(i,L,minn){
    		a[i] += w;
    	}
    	Reset(block[L]);
    	if(block[L] != block[R]){
    		R(i, (block[R] - 1) * blockSize + 1, R){
    			a[i] += w;
    		}
    		Reset(block[R]);
    	}
    	R(i,block[L] + 1, block[R] - 1){
    		tag[i] += w;
    	}
    }
    inline int Query(int L, int R, int w){
    	int sum = 0;
    	int minn = Min(R, block[L] * blockSize);
    	R(i,L,minn){
    		sum += ((a[i] + tag[block[i]]) >= w);
    	}
    	if(block[L] != block[R]){
    		R(i, (block[R] - 1) * blockSize + 1, R){
    			sum += ((a[i] + tag[block[i]]) >= w);
    		}
    	}
    	R(i,block[L] + 1, block[R] - 1){
    		sum += blockSize - (lower_bound(V[i].begin(), V[i].end(), w - tag[i]) - V[i].begin());
    	}
    	
    	return sum;
    }
    
    int main(){
    	int m;
    	io >> n >> m;
    	blockSize = sqrt(n);
    	R(i,1,n){
    		io >> a[i];
    		block[i] = (i - 1) / blockSize + 1;
    		V[block[i]].push_back(a[i]);
    	}
    	
    	int cnt = n / blockSize + 1;
    	R(i,1,cnt)
    		sort(V[i].begin(), V[i].end());
    	
    	while(m--){
    		char opt;
    		int L, R, w;
    		cin >> opt;
    		io >> L >> R >> w;
    		if(opt == 'M'){
    			Updata(L, R, w);
    		}
    		else{
    			printf("%d
    ", Query(L, R, w));
    		}
    	}
    
    	return 0;
    }
    

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  • 原文地址:https://www.cnblogs.com/bingoyes/p/11227353.html
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