HDU 3480 DP 斜率优化 Division

把n个数分成m段，每段的值为(MAX - MIN)²，求所能划分得到的最小值。

依然是先从小到大排个序，定义状态d(j, i)表示把前i个数划分成j段，所得到的最小值，则有状态转移方程：

d(j, i) = min { d(j-1, k) + (a_i - a_k+1)² | 0 ≤ k < i }

设 l < k < i，且由k转移得到的状态比由l转移得到的状态更优。

有不等式：

整理成斜率形式：

后面的就可以相当于套模板了，不过这里要用滚动数组优化一下空间。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 10000 + 10;
const int maxm = 5000 + 10;
const int INF = 0x3f3f3f3f;

int n, m;

int a[maxn];
int d[2][maxn];

int head, tail;
int Q[maxn];

int cur;

int inline Y(int x) { return d[cur^1][x] + a[x+1] * a[x+1]; }

int inline DY(int p, int q) { return Y(q) - Y(p); }

int inline DX(int p, int q) { return a[q+1] - a[p+1]; }

int main()
{
    freopen("in.txt", "r", stdin);

    int T; scanf("%d", &T);
    for(int kase = 1; kase <= T; kase++)
    {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) scanf("%d", a + i);
        sort(a + 1, a + 1 + n);

        memset(d[0], 0x3f, sizeof(d[0]));
        d[0][0] = 0;
        cur = 0;
        for(int i = 1; i <= m; i++)
        {
            cur ^= 1;
            head = tail = 0;
            Q[tail++] = 0;
            for(int j = 1; j <= n; j++)
            {
                while(head + 1 < tail && DY(Q[head], Q[head+1]) <= DX(Q[head], Q[head+1]) * 2 * a[j]) head++;
                while(head + 1 < tail && DY(Q[tail-1], j) * DX(Q[tail-2], Q[tail-1]) <= DY(Q[tail-2], Q[tail-1]) * DX(Q[tail-1], j)) tail--;
                Q[tail++] = j;
                d[cur][j] = d[cur^1][Q[head]] + (a[j]-a[Q[head]+1]) * (a[j]-a[Q[head]+1]);
            }
        }
        printf("Case %d: %d
", kase, d[cur][n]);
    }

    return 0;
}

下面是四边形不等式优化的代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 10000 + 10;
const int maxm = 5000 + 10;
const int INF = 0x3f3f3f3f;

int n, m;

int a[maxn];
int d[maxm][maxn], s[maxm][maxn];

int main()
{
    int T; scanf("%d", &T);
    for(int kase = 1; kase <= T; kase++)
    {
        scanf("%d%d", &n, &m);

        for(int i = 1; i <= n; i++) scanf("%d", a + i);
        sort(a + 1, a + 1 + n);

        memset(s, 0, sizeof(s));
        for(int i = 1; i <= m; i++)
        {
            int j;
            for(j = 1; j <= i; j++) d[i][j] = 0;
            for(; j <= n; j++) d[i][j] = INF;
        }

        for(int i = 1; i <= n; i++)
        {
            s[1][i] = 0;
            d[1][i] = (a[i] - a[1]) * (a[i] - a[1]);
        }

        for(int i = 2; i <= m; i++)
        {
            s[i][n+1] = n;
            for(int j = n; j > i; j--)
            {
                for(int k = s[i-1][j]; k <= s[i][j+1]; k++)
                {
                    int t = d[i-1][k] + (a[j] - a[k+1]) * (a[j] - a[k+1]);
                    if(t < d[i][j])
                    {
                        d[i][j] = t;
                        s[i][j] = k;
                    }
                }
            }
        }

        printf("Case %d: %d
", kase, d[m][n]);
    }

    return 0;
}