二次联通门 : LibreOJ #2016. 「SCOI2016」美味
/* LibreOJ #2016. 「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\n", Tree.Query (0, _L, b, x, root[l - 1], root[r])); } return 0; } int ZlycerQan = Main (); int main (int argc, char *argv[]) {;}
时间: 2024-10-25 21:23:32