"POJ" 1160 Post Office

http://poj.org/problem?id=1160

Test instructions: There are N cities on a straight line, and there is a post office in the city of P, asking how to build a P post office to make each city to the nearest post office and the smallest. (n<=300, P<=30&p<=n)

#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include < iostream>using namespace Std;const int n=1005;typedef long long ll;ll sum[n], D[n][n];int N, p, X[n], s[n][n];//t= (k+j+ 1)/2, W (k, j) = (2*t-j-k) *x[t]-sum (k, t-1) +sum (t+1, J) inline ll W (int i, int j) {int t= (i+j+1) >>1; return (LL) ((t< <1)-i-j) *x[t]-sum[t-1]+sum[i-1]+sum[j]-sum[t]; }int Main () {while (~scanf ("%d%d", &n, &p)) {for (int i=1; i<=n; ++i) scanf ("%d", &x[i]); sort (x+1, x+1+n); fo R (int i=1; i<=n; ++i) sum[i]=sum[i-1]+x[i];for (int i=2; i<=n; ++i) d[1][i]=w (1, i), s[1][i]=1;for (int i=2; i<=p; ++i) {s[i][n+1]=n;for (int j=n; j>=1;--j) {int l=s[i-1][j], r=s[i][j+1], &pos=s[i][j]; ll &now=d[i][j];now=~ 0ull>>1;for (int k=l; k<=r; ++k) {ll t=d[i-1][k-1]+w (k, j); if (now>=t) now=t, Pos=k;}}} printf ("%lld\n", D[p][n]);} return 0;}

　　

Mom, I didn't think of the equation. Qaq really is SB = =

Start from small to big sort once again $x$.

Set $d (i, j) $ to indicate the minimum distance of $i$ post offices currently in the former $j$ cities:

$ $d (i, J) =min \{D (i-1, k-1) +w (k, j) \}, i<j$$

where $w (i, j) $ represents the cost of $i to j$ to build a post office. Apparently the post office was built on the median = =, making $t= \lfloor \frac{i+j+1}{2} \rfloor$, apparently $w (i, j) = (t-i) x[t]-sum (i, t-1) + (J-T) x[t]-sum (T+1, J) $

Find and quadrilateral inequality is too far away ah = = (although the subject of violence can be over = =)

We tried to re-prove ...

1, $w (i, j) $ obviously satisfies the interval monotonicity (in fact, this is completely less than the bottom of the below = =)

2, first prove $w (I, j) $ of the quadrilateral inequality = =, that is, prove $w (i, j+1)-W (i, j) $ about $i$ monotonically decreasing

$$
\begin{align}
W (i, j+1)-W (i, J)
& = \left ((t-i) x[t]-sum (i, t-1) + (J+1-T) x[t]-sum (t+1, j+1) \right)-\left ((t-i) x[t]-sum (i, t-1) + (J-T) x[t]-sum (t+ 1, j) \right) \ \
& = \left ((j+1-t) x[t]-sum (t+1, j+1) \right)-\left ((j-t) X[t]-sum (t+1, j) \right) \ \
& = X[t]-x[j+1]
\end{align}
$$

Obviously $t < j+1$ is $x[t]-x[j+1] \le 0$ so $w$ satisfies the quadrilateral inequality = =

There is no need to prove that the $d$ also satisfies the quadrilateral = = because of the particularity of the topic, we directly prove the interval monotonicity of the decision $s$ =

Set $d_{k} (i, j) = d (i-1, k-1) + W (k, J) $

Set $k \le K ' \le J < j+1$ by quadrilateral inequalities

$ $w (k, J) +w (K ', j+1) \le W (k, j+1) + W (k ', j) $$

Both sides plus $d (i-1, k-1) +d (i-1, K '-1) $, into

$ $d _{k} (i, J) +d_{k '} (i, j+1) \le d_{k} (i, j+1) + d_{k '} (i, j) $$

That

$ $d _{k} (i, j)-D_{k '} (i, J) \le D_{k} (i, j+1)-d_{k '} (i, j+1) $$

So to all $d_{k '} (i, J) \le D_{k} (i, j) \rightarrow D_{k '} (i, j+1) \le d_{k} (i, j+1) $

So obviously, $s (I, J) \le s (i, j+1) $. The same can be $s (i, j+1) \le s (i+1, j+1) $

So get $s (i, j-1) \le s (i, J) \le S (i+1, j) $ spicy!

But found! Mom, it's not good to move in the same length! Because $d (i, i+1) $ is a good trouble to calculate! We decided to deform the inequality!

Due to $s (I, J) \le S (i+1, J) \rightarrow S (i-1, J) \le S (i, j) $, $s (i, j-1) \le s (i, J) \rightarrow S (i, J) \le s (i, j+1) $.

And then you get the

$ $s (i-1, J) \le S (i, J) \le (i, j+1) $$

This takes $i$ as the stage and then pushes the $j$ to get spicy! So it's so hot.

"POJ" 1160 Post Office