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  • BZOJ1176 [Balkan2007]Mokia(CDQ)

    CDQ裸题,(x), (y), (tim)三维偏序

    #include <cstdio>
    #include <iostream>
    #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 Fill(a,b) memset(a, b, sizeof(a))
    #define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
    #define ll long long
    #define u32 unsigned int
    #define u64 unsigned long long
     
    #define ON_DEBUGG
     
    #ifdef ON_DEBUGG
     
    #define D_e_Line printf("
    ----------
    ")
    #define D_e(x) cout << (#x) << " : " << x << endl
    #define Pause() system("pause")
    #define FileOpen() freopen("in.txt", "r", stdin)
    #define FileSave() freopen("out.txt", "w", stdout)
    #include <ctime>
    #define TIME() fprintf(stderr, "
    time: %.3fms
    ", clock() * 1000.0 / CLOCKS_PER_SEC)
    
    #else
     
    #define D_e_Line ;
    #define D_e(x) ;
    #define Pause() ;
    #define FileOpen() ;
    #define FileSave() ;
    #define TIME() ;
    //char buf[1 << 21], *p1 = buf, *p2 = buf;
    //#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
     
    #endif
     
    using namespace std;
    struct ios{
        template<typename ATP>inline ios& operator >> (ATP &x){
            x = 0; int f = 1; char ch;
            for(ch = getchar(); ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;
            while(ch >= '0' && ch <= '9') x = x * 10 + (ch ^ '0'), ch = getchar();
            x *= f;
            return *this;
        }
    }io;
     
    template<typename ATP>inline ATP Max(ATP a, ATP b){
        return a > b ? a : b;
    }
    template<typename ATP>inline ATP Min(ATP a, ATP b){
        return a < b ? a : b;
    }
    template<typename ATP>inline ATP Abs(ATP a){
        return a < 0 ? -a : a;
    }
    
    const int N= 2000007;
    struct Ques{
    	int pos, tim, x, y, val, tag;
    	bool operator < (const Ques &com) const{
    		if(x != com.x) return x < com.x;
    		if(y != com.y) return y < com.y;
    		return tag < com.tag;
    	}
    }q[N], tmp[N];
    int t[N];
    int n;
    inline void Updata(int x, int w){
    	for(; x <= n; x += x & -x) t[x] += w;
    }
    inline int Query(int x){
    	int s = 0;
    	for(; x; x -= x & -x) s += t[x];
    	return s;
    }
    int ans[N];
    inline void CDQ(int l, int r){
    	if(l == r) return;
    	int mid = (l + r) >> 1;
    	R(i,l,r){
    		if(q[i].tim <= mid && q[i].tag == 0){
    			Updata(q[i].y, q[i].val);
    		}
    		if(q[i].tim > mid && q[i].tag == 1){
    			ans[q[i].pos] += q[i].val * Query(q[i].y);
    		} 
    	}
    	R(i,l,r){
    		if(q[i].tim <= mid && q[i].tag == 0){
    			Updata(q[i].y, -q[i].val);
    		}
    	}
    	int j = l, k = mid + 1;
    	R(i,l,r){
    		if(q[i].tim <= mid)
    			tmp[j++] = q[i];
    		else
    			tmp[k++] = q[i];
    	}
    	R(i,l,r){
    		q[i] = tmp[i];
    	}
    	CDQ(l, mid), CDQ(mid + 1, r);
    }
    int tot;
    int main(){
    	freopen("1176.in", "r",  stdin);
    	freopen("1176.out", "w",  stdout);
    	int S;
    	io >> S >> n;
    	while(1){
    		int opt;
    		io >> opt;
    		if(opt == 3) break;
    		if(opt == 1){
    			++tot;
    			io >> q[tot].x >> q[tot].y >> q[tot].val;
    			q[tot].tim = tot;
    		}
    		else{
    			int X1, X2, Y1, Y2;
    			io >> X1 >> Y1 >> X2 >> Y2;
    			int pos = ++ans[0]; // pos is the order of the querys
    			ans[ans[0]] = S * (X2 - X1 + 1) * (Y2 - Y1 + 1);
    			q[++tot] = (Ques){ pos, tot, X1 - 1, Y1 - 1, 1, 1};
    			q[++tot] = (Ques){ pos, tot, X2, Y2, 1, 1};
    			q[++tot] = (Ques){ pos, tot, X1 - 1, Y2, -1, 1};
    			q[++tot] = (Ques){ pos, tot, X2, Y1 - 1, -1, 1};
    		}
    	}
    	sort(q + 1, q + tot + 1);
    	CDQ(1, tot);
    	R(i,1,ans[0]){
    		printf("%d
    ", ans[i]);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/bingoyes/p/11618921.html
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