Try to find out the wrong in the test
题目背景:
分析:线段树优化DP + 复杂度分析
终于不用打公式了,心累······
首先我们只考虑d的限制,在满足d限制的情况下,对于一个i,可行的上一个区间的端点,一定是一段连续的区间,并且,区间右端点是i,那么我们只需要预处理出left[i]表示最小的可以用于更新i的位置,显然left[i]是单增的,我们直接用一个单调队列就可以维护了。
对于c限制并不是很好处理,于是我们考虑分治,对于一段区间[l, r],我们找出其中c值最大的位置p,递归处理[l, p - 1], [p, r],单独处理最后一个区间包含p的情况。
首先比较显然的是,如果我们要求最后一个转移区间包含p,那么可以从[l, p - 1]转移的i,一定是属于[max(p, l + c[p]), r]的,然后我们来分类讨论一下。
1、left[i] < l且i <= p - 1 + c
显然满足条件的i是[l, i - c], 并且i和i + 1的可行区间只相差一个i - c + 1这个位置的贡献,那么我们可以对于第一个位置在线段树中查询,然后后面的所有直接O(1)加上贡献即可
2、left[i] < l且i > p - 1 + c
显然,满足这个条件的i的可行转移范围全部都是[l, p - 1],那么我们只需要二分出满足这个条件的区间,然后查询[l, p - 1]的答案,之后直接给可行区间打标记即可。
3、l <= left[i] < p
这一段的left[i]都不能确定,所以只能直接在线段树里面查询
4、left[i] >= p
显然无法转移。
乍一看这不是n2logn的吗,真的可能过?
现在我们来分析一下复杂度
首先,对于情况1,一定首先有一个log,接下来考虑会扫过多少数,显然,满足情况1的区间是[max(l + c, p), min(r, p + c - 1)],显然区间长度是小于min(p - l + 1, r - p),相当于取被p分割的区间当中长度较小的一个。显然前面的log,一共有n层,是nlogn,后面的区间长度可以通过类似于启发式合并的思想分析,也是nlogn的。
对于情况2,显然每层只有一个,所以是nlogn
对于情况3,我们可以发现,这种情况当且仅当,改点属于区间的右边部分,但是left[i],是在左边部分,那么显然这种情况对于每一个点,有且仅会出现一次,所以是nlogn的
所以,总的复杂度就是O(nlogn)的,当分治到l == r时,直接更新答案,然后更新到上层区间即可。
Source:
/* created by scarlyw */ #include <cstdio> #include <string> #include <algorithm> #include <cstring> #include <iostream> #include <cmath> #include <cctype> #include <vector> #include <set> #include <queue> inline char read() { static const int IN_LEN = 1024 * 1024; static char buf[IN_LEN], *s, *t; if (s == t) { t = (s = buf) + fread(buf, 1, IN_LEN, stdin); if (s == t) return -1; } return *s++; } ///* template<class T> inline void R(T &x) { static char c; static bool iosig; for (c = read(), iosig = false; !isdigit(c); c = read()) { if (c == -1) return ; if (c == '-') iosig = true; } for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0'); if (iosig) x = -x; } //*/ const int OUT_LEN = 1024 * 1024; char obuf[OUT_LEN], *oh = obuf; inline void write_char(char c) { if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf; *oh++ = c; } template<class T> inline void W(T x) { static int buf[30], cnt; if (x == 0) write_char('0'); else { if (x < 0) write_char('-'), x = -x; for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48; while (cnt) write_char(buf[cnt--]); } } inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); } /* template<class T> inline void R(T &x) { static char c; static bool iosig; for (c = getchar(), iosig = false; !isdigit(c); c = getchar()) if (c == '-') iosig = true; for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0'); if (iosig) x = -x; } //*/ const int MAXN = 1000000 + 10; const int mod = 1000000000 + 7; int n; int c[MAXN], d[MAXN], left[MAXN], f[MAXN], g[MAXN]; inline void add(int &x, int t) { x += t, (x >= mod) ? (x -= mod) : (0); } struct node { int c_id, f, g, tag_f, tag_g; } tree[MAXN << 2 | 1]; inline void update(int &f, int &g, int cur_f, int cur_g) { if (cur_f > f) f = cur_f, g = cur_g; else if (f == cur_f) add(g, cur_g); } inline void update(int k) { node lc = tree[k << 1], rc = tree[k << 1 | 1]; tree[k].c_id = ((c[lc.c_id] > c[rc.c_id]) ? lc.c_id : rc.c_id); if (lc.f > rc.f) tree[k].f = lc.f, tree[k].g = lc.g; else if (lc.f == rc.f) tree[k].f = lc.f, tree[k].g = (lc.g + rc.g) % mod; else tree[k].f = rc.f, tree[k].g = rc.g; } inline void modify(int k, int f, int g) { update(tree[k].f, tree[k].g, f, g); update(tree[k].tag_f, tree[k].tag_g, f, g); } inline void push_down(int k) { if (tree[k].tag_f > -1) { modify(k << 1, tree[k].tag_f, tree[k].tag_g); modify(k << 1 | 1, tree[k].tag_f, tree[k].tag_g); tree[k].tag_f = -1, tree[k].tag_g = 0; } } inline void build(int k, int l, int r) { tree[k].tag_f = -1, tree[k].tag_g = 0; if (l == r) { tree[k].c_id = l, tree[k].f = -1, tree[k].g = 0; return ; } int mid = l + r >> 1; build(k << 1, l, mid), build(k << 1 | 1, mid + 1, r); update(k); } inline void modify(int k, int l, int r, int ql, int qr, int f, int g) { if (ql <= l && r <= qr) return modify(k, f, g); int mid = l + r >> 1; if (ql <= mid) modify(k << 1, l, mid, ql, qr, f, g); if (qr > mid) modify(k << 1 | 1, mid + 1, r, ql, qr, f, g); } inline void modify(int k, int l, int r, int pos, int f, int g) { if (l == r) return modify(k, f, g); int mid = l + r >> 1; push_down(k); if (pos <= mid) modify(k << 1, l, mid, pos, f, g); else modify(k << 1 | 1, mid + 1, r, pos, f, g); } inline void query(int k, int l, int r, int ql, int qr, int &f, int &g) { if (qr < ql) return; if (ql <= l && r <= qr) return update(f, g, tree[k].f, tree[k].g); int mid = l + r >> 1; if (ql <= mid) query(k << 1, l, mid, ql, qr, f, g); if (qr > mid) query(k << 1 | 1, mid + 1, r, ql, qr, f, g); } inline void query(int k, int l, int r, int pos, int &f, int &g) { if (l == r) return update(f, g, tree[k].f, tree[k].g); int mid = l + r >> 1; if (pos <= mid) query(k << 1, l, mid, pos, f, g); else query(k << 1 | 1, mid + 1, r, pos, f, g); update(k); } inline int find(int k, int l, int r, int ql, int qr) { if (ql <= l && r <= qr) return tree[k].c_id; int mid = l + r >> 1; if (qr <= mid) return find(k << 1, l, mid, ql, qr); else if (ql > mid) return find(k << 1 | 1, mid + 1, r, ql, qr); else { int pos1 = find(k << 1, l, mid, ql, qr); int pos2 = find(k << 1 | 1, mid + 1, r, ql, qr); return (c[pos1] > c[pos2]) ? pos1 : pos2; } } inline void read_in() { R(n); for (int i = 1; i <= n; ++i) R(c[i]), R(d[i]); } inline void pre_work() { int cur_pos = 0, head = 1, tail = 0; static int q[MAXN]; for (int i = 1; i <= n; ++i) { while (head <= tail && d[q[tail]] >= d[i]) tail--; q[++tail] = i; int pos; while (head <= tail && q[head] <= cur_pos) head++; pos = d[q[head]]; while (i - pos > cur_pos) { ++cur_pos; while (head <= tail && q[head] <= cur_pos) head++; pos = d[q[head]]; } left[i] = cur_pos; } // for (int i = 1; i <= n; ++i) std::cout << left[i] << " "; // std::cout << '\n'; } inline int binary(int x, int y, int val) { int l = x, r = y + 1; while (l + 1 < r) { int mid = l + r >> 1; (left[mid] < val) ? l = mid : r = mid; } // std::cout << "??" << l << '\n'; return l; } inline void solve(int l, int r, int k) { int start = std::max(k, l + c[k]), end1 = std::min(r, k + c[k] - 1); int cur_f = -1, cur_g = 0, st = start; while (st <= end1) { if (left[st] >= l) break ; if (left[st] >= k) return ; if (st == start) query(1, 0, n, l, st - c[k], cur_f, cur_g); else update(cur_f, cur_g, f[st - c[k]], g[st - c[k]]); if (cur_f != -1) update(f[st], g[st], cur_f + 1, cur_g); ++st; } if (st > r) return ; if (left[st] < l) { int pos = binary(st, r, l); cur_f = -1, cur_g = 0; query(1, 0, n, l, k - 1, cur_f, cur_g); if (cur_f > -1) modify(1, 0, n, st, pos, cur_f + 1, cur_g); st = pos + 1; } while (st <= r) { if (left[st] >= k) return ; cur_f = -1, cur_g = 0; query(1, 0, n, left[st], std::min(k - 1, st - c[k]), cur_f, cur_g); if (cur_f > -1) update(f[st], g[st], cur_f + 1, cur_g); st++; } } inline void solve(int l, int r) { if (r < l) return ; if (l == r) { modify(1, 0, n, l, f[l], g[l]); f[l] = -1, g[l] = 0, query(1, 0, n, l, f[l], g[l]); return ; } int pos = find(1, 0, n, l + 1, r); solve(l, pos - 1), solve(l, r, pos), solve(pos, r); } int main() { freopen("schooldays.in", "r", stdin); freopen("schooldays.out", "w", stdout); read_in(); pre_work(); build(1, 0, n); for (int i = 1; i <= n; ++i) f[i] = -1; g[0] = 1, solve(0, n); // for (int i = 1; i <= n; ++i) std::cout << f[i] << " " << g[i] << '\n'; if (f[n] > -1) std::cout << f[n] << " " << g[n]; else std::cout << "-1"; return 0; }