luogu P7416 [USACO21FEB] No Time to Dry P
阿新 • • 發佈:2021-12-21
https://www.luogu.com.cn/problem/P7416
對於每個數,記\(lst[i]\)表示上一個和\(i\)顏色相同的位置
如果\(lst[i]<l\)顯然\(ans+1\)
如果\(min(lst[i],i)<a[i]\)那麼\(ans+1\)(把\(lst[i]\)設為\(0\)即可)
然後就變成了詢問區間\(lst[i]<l\)的個數
拿主席樹維護一下即可
不對啊,可以離線,直接樹狀陣列即可
我是sg
code:
#include<bits/stdc++.h> #define N 400050 using namespace std; #define ls (a[rt].l) #define rs (a[rt].r) struct A { int l, r, s; } a[N << 5]; void update(int rt) { a[rt].s = a[ls].s + a[rs].s; } int tot; void add(int &rt, int lrt, int l, int r, int x, int o) { rt = ++ tot; a[rt] = a[lrt]; if(l == r) { a[rt].s += o; return ; } int mid = (l + r) >> 1; if(x <= mid) add(ls, a[lrt].l, l, mid, x, o); else add(rs, a[lrt].r, mid + 1, r, x, o); update(rt); } int query(int rt, int l, int r, int L, int R) { if(!rt) return 0; if(L <= l && r <= R) return a[rt].s; int mid = (l + r) >> 1, ret = 0; if(L <= mid) ret = query(ls, l, mid, L, R); if(R > mid) ret += query(rs, mid + 1, r, L, R); return ret; } int mi[N][21]; int get(int l, int r) { int k = log2(r - l + 1); return min(mi[l][k], mi[r - (1 << k) + 1][k]); } int n, m, aa[N], lst[N], root[N]; map<int, int> mp; int main() { // freopen("a.in","r",stdin); // freopen("a.out","w",stdout); scanf("%d%d", &n, &m); for(int i = 1; i <= n; i ++) scanf("%d", &aa[i]), lst[i] = mp[aa[i]], mp[aa[i]] = i, mi[i][0] = aa[i]; for(int j = 1; j <= 19; j ++) for(int i = 1; i + (1 << j) - 1 <= n; i ++) mi[i][j] = min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]); for(int i = 1; i <= n; i ++) if(get(lst[i] + 1, i) < aa[i]) lst[i] = 0; // for(int i = 1; i <= n; i ++) printf("%d ", lst[i]); printf("\n"); for(int i = 1; i <= n; i ++) add(root[i], root[i - 1], 0, n, lst[i], 1); // for(int i = 1; i <= n; i ++) printf("%d ", root[i]); printf(" %d\n", n); while(m --) { int l, r; scanf("%d%d", &l, &r); // printf("%d %d ", query(root[r], 0, n, 0, l - 1), query(root[l - 1], 0, n, 0, l - 1)); printf("%d\n", query(root[r], 0, n, 0, l - 1) - query(root[l - 1], 0, n, 0, l - 1)); } return 0; }