Description
有n个人,要分成m个班,每个班至少1人。每班贡献为((max-min)^2),求最小贡献值。
(1<=N<=10000,1<=M<=1000,1<=Ai<=1000000)
Solution
很容易想到(DP)。
由于我们要使贡献最小,所以我们不妨将它排序,然后每次取连续的一段,这样可以保证值最小。
这题由于要刚好分成(m)个班,所以我们必须要多设一维(DP)。
设(f[i][j])表示到第(i)个人,往前分了(j)个班的最小贡献。
很容易得到转移方程:
[f[i][j] = min(f[k][j - 1] + (a[i] - a[k + 1])^2)
]
时间为(O(n^2m)),过不了,所以我们考虑优化。
斜率优化即可。
Code
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 10010
#define M 1010
#define mem(x, a) memset(x, a, sizeof x)
#define fo(x, a, b) for (int x = a; x <= b; x++)
using namespace std;
int n, m, a[N], f[N][M], g[N], l = 1, len = 0;
inline int read()
{
int x = 0; char c = getchar();
while (c < '0' || c > '9') c = getchar();
while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
return x;
}
int sqr(int x) {return x * x;}
int left(int x, int y, int k) {return f[x][k] - f[y][k] + sqr(a[x + 1]) - sqr(a[y + 1]);}
int right(int x, int y) {return 2ll * (a[x + 1] - a[y + 1]);}
int main()
{
freopen("queue.in", "r", stdin);
freopen("queue.out", "w", stdout);
n = read(), m = read();
fo(i, 1, n) a[i] = read();
sort(a + 1, a + n + 1);
mem(f, 10);
fo(i, 1, n) f[i][1] = sqr(a[i] - a[1]);
fo(j, 2, m)
{
f[j][j] = 0;
g[l = 1] = j - 1, g[len = 2] = j;
fo(i, j + 1, n)
{
while (l < len && left(g[l], g[l + 1], j - 1) > right(g[l], g[l + 1]) * a[i]) l++;
f[i][j] = f[g[l]][j - 1] + sqr(a[i] - a[g[l] + 1]);
while (l < len && left(g[len - 1], g[len], j - 1) * right(g[len], i) >= left(g[len], i, j - 1) * right(g[len - 1], g[len])) len--;
g[++len] = i;
}
}
printf("%d
", f[n][m]);
return 0;
}
2019.8.4 update
这题可以用凸优化来弄。(O(nlogn))
凸优化:对于每一次的选取加一个(K)的代价。
显然,(K)越大分成的组数越小,反之越大。
所以,我们可以二分(K),最后答案便是(f[n]-m*K)
Code
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 10010
#define M 1010
#define mem(x, a) memset(x, a, sizeof x)
#define fo(x, a, b) for (int x = a; x <= b; x++)
using namespace std;
int n, m, a[N], f[N][2], g[N], l = 1, len = 0;
inline int read()
{
int x = 0; char c = getchar();
while (c < '0' || c > '9') c = getchar();
while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
return x;
}
int sqr(int x) {return x * x;}
int left(int x, int y) {return f[x][0] - f[y][0] + sqr(a[x + 1]) - sqr(a[y + 1]);}
int right(int x, int y) {return 2ll * (a[x + 1] - a[y + 1]);}
void check(int x)
{
f[0][0] = 0, f[0][1] = 0;
g[l = len = 1] = 0;
fo(i, 1, n)
{
while (l < len && left(g[l], g[l + 1]) > right(g[l], g[l + 1]) * a[i]) l++;
f[i][0] = f[g[l]][0] + sqr(a[i] - a[g[l] + 1]) + x;
f[i][1] = f[g[l]][1] + 1;
while (l < len && left(g[len - 1], g[len]) * right(g[len], i) >= left(g[len], i) * right(g[len - 1], g[len])) len--;
g[++len] = i;
}
}
int main()
{
freopen("queue.in", "r", stdin);
freopen("queue.out", "w", stdout);
n = read(), m = read();
fo(i, 1, n) a[i] = read();
sort(a + 1, a + n + 1);
int l = 0, r = 1000000, mid, K;
while (l <= r)
{
mid = l + r >> 1;
check(mid);
if (f[n][1] == m) break;
if (f[n][1] < m) r = mid - 1;
else l = mid + 1;
}
printf("%d
", f[n][0] - mid * m);
return 0;
}