• [洛谷P3332][ZJOI2013]K大数查询


    题目大意:有$n$个位置,$m$个操作。操作有两种:

    1. $1;l;r;x:$在区间$[l,r]$每个位置加上一个数$x$
    2. $2;l;r;k:$询问$[l,r]$中第$k$大的数是多少。

    题解:树套树,权值线段树套位置线段树,要标记永久化,不然会$TLE$

    卡点:没有标记永久化,$TLE$,然后处理$tag$部分写错

    C++ Code:

    #include <cstdio>
    #include <algorithm>
    #include <cctype>
    namespace __IO {
    	namespace R {
    		int x, ch, f;
    		inline int readsign() {
    			f = 1;
    			while (isspace(ch = getchar()));
    			if (ch == '-') f = -1;
    			for (x = ch & 15; isdigit(ch = getchar()); ) x = x * 10 + (ch & 15);
    			return x * f;
    		}
    		inline int read() {
    			while (isspace(ch = getchar()));
    			for (x = ch & 15; isdigit(ch = getchar()); ) x = x * 10 + (ch & 15);
    			return x;
    		}
    
    		long long X;
    		inline long long readll() {
    			while (isspace(ch = getchar()));
    			for (X = ch & 15; isdigit(ch = getchar()); ) X = X * 10 + (ch & 15);
    			return X;
    		}
    	}
    }
    using __IO::R::read;
    using __IO::R::readsign;
    using __IO::R::readll;
    
    #define maxn 50010
    
    int n, m;
    
    namespace SgT2 {
    #define N ((maxn << 3) * 50)
    	int lc[N], rc[N], tg[N], idx;
    	int L, R;
    	long long V[N];
    
    	void __insert(int &rt, const int l, const int r) {
    		if (!rt) rt = ++idx;
    		V[rt] += std::min(R, r) - std::max(L, l) + 1;
    		if (L <= l && R >= r) {
    			tg[rt]++;
    			return ;
    		}
    		int mid = l + r >> 1;
    		if (L <= mid) __insert(lc[rt], l, mid);
    		if (R > mid) __insert(rc[rt], mid + 1, r);
    	}
    	void insert(int &rt, int __L, int __R) {
    		L = __L, R = __R;
    		__insert(rt, 1, n);
    	}
    
    	long long __query(const int rt, const int l, const int r) {
    		if (!rt || (L <= l && R >= r)) return V[rt];
    		int mid = l + r >> 1;
    		long long res = static_cast<long long> (std::min(R, r) - std::max(L, l) + 1) * tg[rt];
    		if (L <= mid) res += __query(lc[rt], l, mid);
    		if (R > mid) res += __query(rc[rt], mid + 1, r);
    		return res;
    	}
    	long long query(int rt, int __L, int __R) {
    		L = __L, R = __R;
    		return __query(rt, 1, n);
    	}
    #undef N
    }
    
    namespace SgT {
    #define N (maxn << 3)
    	const int maxl = -50000, maxr = 50000;
    	int root[N];
    	int L, R, num;
    	void __insert(const int rt, const int l, const int r) {
    		SgT2::insert(root[rt], L, R);
    		if (l == r) return ;
    		int mid = l + r >> 1;
    		if (num <= mid) __insert(rt << 1, l, mid);
    		else __insert(rt << 1 | 1, mid + 1, r);
    	}
    	void insert(int __L, int __R, int __num) {
    		L = __L, R = __R, num = __num;
    		__insert(1, maxl, maxr);
    	}
    
    	long long pos;
    	int __query(const int rt, const int l, const int r) {
    		if (l == r) return l;
    		int mid = l + r >> 1;
    		long long tmp = SgT2::query(root[rt << 1 | 1], L, R);
    		if (pos <= tmp) return __query(rt << 1 | 1, mid + 1, r);
    		else {
    			pos -= tmp;
    			return __query(rt << 1, l, mid);
    		}
    	}
    	int query(int __L, int __R, long long __pos) {
    		L = __L, R = __R, pos = __pos;
    		return __query(1, maxl, maxr);
    	}
    #undef N
    }
    using SgT::insert;
    using SgT::query;
    
    int main() {
    	n = read(), m = read();
    	while (m --> 0) {
    		int op = read(), l = read(), r = read();
    		if (op == 1) {
    			int c = readsign();
    			insert(l, r, c);
    		} else {
    			long long c = readll();
    			printf("%d
    ", query(l, r, c));
    		}
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Memory-of-winter/p/10104954.html
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