JZOJ 5646. 【NOI2018模拟4.12】染色游戏

最新推荐文章于 2021-03-16 20:26:10 发布

原创最新推荐文章于 2021-03-16 20:26:10 发布 · 361 阅读

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本文介绍了一种利用斜率优化解决特定数学问题的方法，并结合CDQ分治策略来处理带有额外约束条件的问题。通过具体代码实现展示了如何有效地进行优化计算。

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题目

这里写图片描述

分析

很显然的斜率优化，如果我们考虑忽略a那个限制的话，对于a的限制我们可以考虑cdq分治来做

代码

#include <bits/stdc++.h>

const int N = 1e6 + 10;

typedef long long ll;

int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar();}
    while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
    return x * f;
}

struct Data
{
    int v,id;
}a[N];

struct Note
{
    int x,y;
}t[25][N];

void pre(int l,int r,int d)
{
    if (l == r)
    {
        t[d][l].y = a[l].id;
        t[d][l].x = l; 
        return;
    }
    int mid = (l + r) >> 1, i, j, k;
    pre(l, mid, d + 1);
    pre(mid + 1, r, d + 1);
    for (i = k = l, j = mid + 1; i <= mid && j <= r; k++)
        if (t[d + 1][i].y < t[d + 1][j].y) 
            t[d][k] = t[d + 1][i++];
        else t[d][k] = t[d + 1][j++];
    for (; i <= mid; t[d][k++] = t[d + 1][i++]);
    for (; j <= r; t[d][k++] = t[d + 1][j++]);
}

bool operator < (Data a,Data b)
{
    return a.v < b.v || a.v == b.v && a.id < b.id;
}

double Get(double k1,double b1,double k2,double b2)
{
    return (b2 - b1) / (k1 - k2);
}

ll f[N];
ll x[N],y[N];

int Q[N];

void solve(int l,int r,int d)
{
    if (l == r)
    {
        x[l] = a[l].id, y[l] = f[l] - 1ll * x[l] * x[l];
        return ;
    }
    int mid = (l + r) >> 1;
    solve(l, mid, d + 1);
    int h = 1, tail = 0;
    for (int i = l; i <= r; i++)
    {
        int j = t[d][i].x;
        if (j <= mid)
        {
            for (; h < tail && Get(x[j], y[j], x[Q[tail - 1]], y[Q[tail - 1]]) <= Get(x[Q[tail - 1]], y[Q[tail - 1]], x[Q[tail]], y[Q[tail]]); tail--);
            Q[++tail] = j;
        }
        else
        {
            for (; h < tail && Get(x[Q[h]], y[Q[h]], x[Q[h + 1]], y[Q[h + 1]]) <= a[j].id; h++);
            if (h <= tail) 
                f[j] = std::max(f[j], a[j].v + 1ll * a[j].id * x[Q[h]] + y[Q[h]]);
        }
    }
    solve(mid + 1, r, d + 1);
}

int main()
{
    freopen("paint.in","r",stdin); 
    freopen("paint.out","w",stdout);
    int n = read();
    for (int i = 1; i <= n; i++)
        a[i].v = read(), a[i].id = i;
    std::sort(a + 1, a + n + 1);
    pre(1,n,0);
    for (int i = 1; i <= n; i++)
        f[i] = a[i].v;
    solve(1,n,0);
    ll ans = 0;
    for (int i = 1; i <= n; i++)
        ans = std::max(ans, f[i] + 1ll * (n + 1 - a[i].id) * (a[i].id));
    ans -= 1ll * n * (n + 1) / 2ll;
    printf("%lld\n", ans);
}