题目描述
四边形不等式优化,看决策候选集形式直接猜是四边形不等式优化,中间再用前缀和加二分优化一下,可以做到O(NPlogN)
时间复杂度 O(NPlogN)
参考文献
C++ 代码
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 310, M = 50, INF = 0x3f3f3f3f;
int f[N][M];
int opt[N][N];
int s[N];
int pos[N];
int n, m;
int val(int k, int i ,int j)
{
int res = f[k][j - 1];
int l = k, r = i;
while(l < r)
{
int mid = (l + r) / 2;
if(pos[i] - pos[mid] <= pos[mid] - pos[k]) r = mid;
else l = mid + 1;
}
res += pos[i] * (i - r + 1) - s[i] + s[r - 1];
res += s[r - 1] - s[k - 1] - (r - k) * pos[k];
return res;
}
int main()
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++ i) scanf("%d", &pos[i]);
sort(pos + 1, pos + n + 1);
for(int i = 1; i <= n; ++ i) s[i] = s[i - 1] + pos[i];
memset(f, 0x3f, sizeof f);
f[1][1] = 0;
for(int i = 2; i <= n; ++ i) f[i][1] = (i - 1) * pos[i] - s[i - 1];
for(int i = 2; i <= n; ++ i)
for(int j = 2; j <= m; ++ j)
{
if(j > n) continue;
if(i == j) opt[i][j] = i - 1, f[i][j] = 0;
else{
for(int k = opt[i - 1][j]; k < i; ++ k)
{
if(val(k, i, j) < f[i][j])
{
f[i][j] = val(k, i, j);
opt[i][j] = k;
}
}
}
}
int ans = INF;
for(int i = m; i <= n; ++ i) ans = min(ans, f[i][m] + s[n] - s[i] - (n - i) * pos[i]);
printf("%d", ans);
return 0;
}
内层的$k$均摊是常数级别吧,整体上$O(NPlogN)$
这个复杂度应该是 $O(N^2P\log N)$ 级别的吧