BZOJ4709: [Jsoi2011]柠檬(斜率优化)

题意

题目链接

sol

结论：每次选择的区间一定满足首位元素相同。。

仔细想想其实挺显然的，如果不相同可以删掉多着的元素，对答案的贡献是相同的

那么设\(f[i]\)表示到第\(i\)个位置的最大价值，\(s[i]\)表示到\(i\)位置，\(a[i]\)的出现次数，转移方程为

\[f[i] = max(f_{j - 1} + a[i] * (s[i] - s[j] +1)^2)\]

满足\(a[i] = a[j]\)

看起来好像是可以斜率优化的样子，不过存在另外一种解释。。

具体看吧

感觉自己的斜率优化学的狠不到家啊，，有空补补qwq

#include<bits/stdc++.h>
#define chmax(a, b) (a = (a > b ? a : b))
#define chmin(a, b) (a = (a < b ? a : b))
#define ll long long
//#define int long long 
using namespace std;
const int maxn = 1e6 + 10;
inline int read() {
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, a[maxn], s[maxn], cnt[maxn]; ll f[maxn];//s[i] i位置的数第几次出?
vector<int> v[maxn];
ll calc(int pos, ll val) {
    return f[pos - 1] + 1ll * a[pos] * val * val;
}
int lower(int x, int y) {
    int l = 1, r = n, ans = n + 1;
    while(l <= r) {
        int mid = l + r >> 1;
        if(calc(x, mid - s[x] + 1) >= calc(y, mid - s[y] + 1)) r = mid - 1, ans = mid;
        else l = mid + 1;
    }
    return ans;
}
main() {
    n = read();
    for(int i = 1, siz, s; i <= n; i++) {
        s[i] = ++cnt[s = a[i] = read()];
        while((((siz = v[s].size()) >= 2) && (lower(v[s][siz - 2], v[s][siz - 1]) <= lower(v[s][siz - 1], i)))) v[s].pop_back();
        v[s].push_back(i);
        while(((siz = v[s].size()) >= 2) && (lower(v[s][siz - 2], v[s][siz - 1]) <= s[i]))v[s].pop_back();
        f[i] = calc(v[s][v[s].size() - 1], s[i] - s[v[s][v[s].size() - 1]] + 1);
    }
    cout << f[n];
}