「POI2015」KIN

Portal
Luogu

Problem-solving ideas

This question is do you want, as long as maintains the largest sub-segment interval and just fine.
And this can be maintained with a segment tree template Click here
for the case of repetition, we can write a precursor same kind of representation and the current position of a position in front of each location.
Then to eliminate the contribution by this.
So they have done?

Details Notes

• Gugu Gu

Reference Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while (!isdigit(c)) f |= (c == '-'), c = getchar();
    while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
    s = f ? -s : s;
}

typedef long long LL;
const int _ = 1000010;

int n, m, f[_], w[_], pre[_], las[_];
struct node { LL sum, L, R, mx; }t[_ << 2];

inline int lc(int rt) { return rt << 1; }

inline int rc(int rt) { return rt << 1 | 1; }

inline void pushup(int rt) {
    t[rt].sum = t[lc(rt)].sum + t[rc(rt)].sum;
    t[rt].L = max(t[lc(rt)].L, t[lc(rt)].sum + t[rc(rt)].L);
    t[rt].R = max(t[rc(rt)].R, t[rc(rt)].sum + t[lc(rt)].R);
    t[rt].mx = max(t[lc(rt)].R + t[rc(rt)].L, max(t[lc(rt)].mx, t[rc(rt)].mx));
}

inline void update(int id, LL v, int rt = 1, int l = 1, int r = n) {
    if (l == r) { t[rt] = (node) { v, v, v, v }; return; }
    int mid = (l + r) >> 1;
    if (id <= mid) update(id, v, lc(rt), l, mid);
    else update(id, v, rc(rt), mid + 1, r);
    pushup(rt);
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.in", "r", stdin);
#endif
    read(n), read(m);
    for (rg int i = 1; i <= n; ++i) read(f[i]);
    for (rg int i = 1; i <= m; ++i) read(w[i]);
    for (rg int i = 1; i <= n; ++i)
        pre[i] = las[f[i]], las[f[i]] = i;
    LL ans = 0;
    for (rg int i = 1; i <= n; ++i) {
        if (pre[i]) update(pre[i], -w[f[i]]);
        if (pre[pre[i]]) update(pre[pre[i]], 0);
        update(i, (LL) w[f[i]]), ans = max(ans, t[1].mx);
    }
    printf("%lld\n", ans);
    return 0;
}

End Sahua \ (qwq \)