Educational Codeforces Round 97 (Rated for Div. 2)
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E. Make It Increasing Educational Codeforces Round 97 (Rated for Div. 2)
题解:
因为b[i]的值是确定,所以知道操作 b[i] 到b[i - 1]的区间, 当 (a[i] - a[j] < i - j (i > j)) 那么可以证明一定无法构成, 所以先判断是否可以构造,如果都满足, 那么对与每段 (b[i - 1])到 (b[i]) 可以先将(a[i] = a[i] - i)
这样做的目的是保证存在 (a[i] > a[j] ,且 i > j) 那么就是且以 (b[i - 1])开始 (b[i]) 结束的最长上升子序列, 可以用二分优化到 n * log(n)也可以用线段树优化。
代码:
#include<bits/stdc++.h>
using namespace std;
const int N = 5e5 + 7;
typedef long long ll;
ll a[N], n, k, b[N];
int rt;
int tree[40 * N], ls[40 * N], rs[40 * N], dp[N];
int top = 1;
#define m (l + r) / 2
void update(int v, int pos, int l, int r, int &node) {
if (node == 0) node = top++;
if (l == r) {
tree[node] = v;
return;
}
if (pos <= m) update(v, pos, l, m, ls[node]);
else update(v, pos, m + 1, r, rs[node]);
tree[node] = max(tree[ls[node]], tree[rs[node]]);
}
int query(int ql, int qr, int l, int r, int node) {
if (ql <= l && qr >= r) {
return tree[node];
}
int ans = 0;
if (ql <= m) ans = max(ans, query(ql, qr, l, m, ls[node]));
if (qr > m) ans = max(ans, query(ql, qr, m + 1, r, rs[node]));
return ans;
}
int work(int l, int r) {
int maxn = n + 2;
for (int i = l; i <= r; i++) {
dp[i] = 0;
}
for (int i = l + 1; i <= r; i++) {
if (a[i] >= a[l]) {
dp[i] = 2;
} else {
dp[i] = -1e8;
}
}
for (int i = l + 1; i <= r; i++) {
dp[i] = max(dp[i], query(1, a[i], 1, maxn, rt) + 1);
if (dp[i] > 1)
update(dp[i], a[i], 1, maxn, rt);
}
for (int i = l; i <= r; i++) {
update(0, a[i], 1, maxn, rt);
}
if (dp[r] < 2) {
return -1;
}
return (r - l + 1) - dp[r];
}
vector<int> g;
int get_id(int x) {
return lower_bound(g.begin(), g.end(), x) - g.begin() + 1;
}
int main() {
ios::sync_with_stdio(0);
cin >> n >> k;
for (int i = 1; i <= n; i++) {
cin >> a[i];
a[i] = a[i] - i;
g.push_back(a[i]);
}
sort(g.begin(), g.end());
g.erase(unique(g.begin(), g.end()), g.end());
for (int i = 1; i <= k; i++) {
cin >> b[i];
}
for (int i = 1; i <= n; i++) {
a[i] = get_id(a[i]);
}
b[0] = 0;
b[k + 1] = n + 1;
a[0] = 0;
a[n + 1] = n + 2;
int ans = 0;
for (int i = 1; i <= k + 1; i++) {
int cnt = work(b[i - 1], b[i]);
if (cnt < 0) {
cout << -1 << endl;
return 0;
}
ans += cnt;
}
cout << ans << endl;
}
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