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) + (ai - ak+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：

#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\n", kase, d[m]
);
}

return 0;
}

代码君