• BZOJ 4571: [Scoi2016]美味


    二次联通门 : BZOJ 4571: [Scoi2016]美味

    /*
        BZOJ 4571: [Scoi2016]美味
        
        dalao们都在说这题如果没有加法balabala就可以用可持久化trie解决了
        然而我连那个也不会啊QAQ
    
        此题用主席树
        从高位到低位贪心
        能填1就填1,也就是查询一段区间有没有某个范围的数
        (然而由乃dalao说可持久化线段树和可持久化trie是一个东西)
    */
    #include <cstdio>
    #include <iostream>
    #include <cstring>
    
    const int BUF = 12312313;
    char Buf[BUF], *buf = Buf;
    
    inline void read (int &now)
    {
        for (now = 0; !isdigit (*buf); ++ buf);
        for (; isdigit (*buf); now = now * 10 + *buf - '0', ++ buf);
    }
    
    const int _L = (1 << 19) - 1;
    #define Max 200010
    int root[Max * 20], c[Max], S;
    inline int max (int a, int b) { return a > b ? a : b; }
    class President_Tree
    {
        private :
    
            int tree[Max * 30][2], size[Max * 30];
            int T_C;
    
        public :
    
            President_Tree () { T_C = 0; }
            void Change (int &now, int last, int l, int r, int key)
            {
                now = ++ T_C;
                memcpy (tree[now], tree[last], sizeof tree[now]);
                size[now] = size[last] + 1;
                if (l == r) return ;
                int Mid = l + r >> 1;
                if (key <= Mid)
                    Change (tree[now][0], tree[last][0], l, Mid, key);
                else Change (tree[now][1], tree[last][1], Mid + 1, r, key);
            }
    
            int Query (int l, int r, int b, int x, int L, int R)
            {
                if (l == r) return b ^ l;
                int Mid = l + r >> 1; -- S;
                if (b & (1 << S))
                {
                    if (Q (L, R    , 0, _L, max (0, l - x), max (0, Mid - x)))
                        return Query (l, Mid, b, x, L, R);
                    else return Query (Mid + 1, r, b, x, L, R);
                }
                else 
                {
                    if (Q (L, R, 0, _L, max (0, Mid + 1 - x), max (0, r - x)))
                        return Query (Mid + 1, r, b, x, L, R);
                    else return Query (l, Mid, b, x, L, R);
                }
            }
    
            int Q (int L, int R, int x, int y, int l, int r)
            {
                if (x == l && y == r) return size[R] - size[L];
                int Mid = x + y >> 1;
                if (r <= Mid) 
                    return Q (tree[L][0], tree[R][0], x, Mid, l, r);
                else if (l > Mid)
                    return Q (tree[L][1], tree[R][1], Mid + 1, y, l, r);
                else 
                    return Q (tree[L][0], tree[R][0], x, Mid, l, Mid) + Q (tree[L][1], tree[R][1], Mid + 1, y, Mid + 1, r);
            }
    };
    President_Tree Tree;
    
    int Main ()
    {
        fread (buf, 1, BUF, stdin);
        int N, M; register int i, j; read (N), read (M);
        int x, b, l, r;
        for (i = 1; i <= N; ++ i) 
            read (c[i]), Tree.Change (root[i], root[i - 1], 0, _L, c[i]);
        
        for (i = 1; i <= M; ++ i)
        {
            read (b), read (x), read (l), read (r); S = 19;
            printf ("%d
    ", Tree.Query (0, _L, b, x, root[l - 1], root[r]));
        }
        return 0;
    }
    int ZlycerQan = Main ();
    int main (int argc, char *argv[]) {;}
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  • 原文地址:https://www.cnblogs.com/ZlycerQan/p/7436239.html
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