5125: [Lydsy1712月赛]小Q的书架 [决策单调性]
作者:互联网
很显然的dp式子
\(dp_i = dp_j + calc(j + 1,i)\)
然后发现这个 \(calc(i-1,j+1) + calc(i,j) \geq calc(i-1,j) + calc(i,j+1)\)
满足四边形不等式,然后就可以决策单调性,区间逆序对搞个莫队就完了。
#include <bits/stdc++.h>
using namespace std;
template <int maxn>
struct BIT {
int c[maxn];
int low(int x) { return x & -x; }
void add(int x, int y) {
for (; x < maxn; x += low(x)) c[x] += y;
}
int qry(int x) {
int ans = 0;
for (; x; x ^= low(x)) ans += c[x];
return ans;
}
};
int n, k;
const int maxn = 4e4 + 44;
BIT<maxn> bit;
int a[maxn], f[maxn], g[maxn];
int l, r, ans = 0;
void modify(int ql, int qr) {
while (l < ql) {
bit.add(a[l], -1);
ans -= bit.qry(a[l++] - 1);
}
while (l > ql) {
bit.add(a[--l], 1);
ans += bit.qry(a[l] - 1);
}
while (r < qr) {
bit.add(a[++r], 1);
ans += bit.qry(n) - bit.qry(a[r]);
}
while (r > qr) {
bit.add(a[r], -1);
ans -= bit.qry(n) - bit.qry(a[r--]);
}
}
void solve(int l, int r, int L, int R) {
if (l > r) return;
int mid = l + r >> 1, p;
for (int i = min(mid - 1, R); i >= L; --i) {
modify(i + 1, mid);
if (g[i] + ans < f[mid]) f[mid] = g[i] + ans, p = i;
}
solve(l, mid - 1, L, p);
solve(mid + 1, r, p, R);
}
signed main() {
scanf("%d %d", &n, &k);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
l = 1, r = 0;
for (int i = 1; i <= n; i++) {
modify(1, i), f[i] = ans;
}
for (int i = 2; i <= k; i++) {
for (int j = 1; j <= n; j++) g[j] = f[j];
for (int j = 1; j <= n; j++) f[j] = 1e9;
solve(1, n, 1, n);
}
printf("%d\n", f[n]);
return 0;
}
标签:月赛,int,qry,mid,maxn,ans,5125,bit,Lydsy1712 来源: https://www.cnblogs.com/Isaunoya/p/12625825.html