HOJ-1005

这题一看就像是动归，但我就是想不出来怎么动归。。。

纠结了n天之后，果断百度，果然是动归

a[i][j]表示前i个饭店如果需要j个depot

c[i][j]表示如果从第i个到第j个饭店都由一个depot负责，的最短距离

于是就有了神奇的动态转移方程

　　a[i][j] = min{a[i][j], a[i-k][j-1]+c[i-k+1][i]} (k = 1, 2, …, i-1)

嗯，然后就益母聊染了

#include<cstdio>
#include<cstdlib>
#include<cstring>

#define min(a,b) (((a)>(b))?(b):(a))

#define MAXN 300

int d[MAXN];//distance between restautants i and headquarter
int a[MAXN][MAXN];//the lowest distance while the first i restautants need j depots
int c[MAXN][MAXN];//the lowest distance while restautant i to restaurant j need 1 depots

int main(void)
{
  int n, k;
  int num = 1;
  scanf("%d%d", &n, &k);
  while(n > 0 && k > 0){
    for(int i = 1; i <= n; ++i) scanf("%d", &d[i]);
    memset(c, 0, sizeof(c));
    for(int i = 1; i <= n; ++i){
      for(int j = i+1; j <= n; ++j)
    for(int k = i, p = (i+j)/2; k <= j; ++k)
      c[i][j] += abs(d[k] - d[p]);
    }
    for(int i = 0; i <= n; ++i)
      for(int j = 0; j <= n; ++j)
    a[i][j] = 0xFFFFFFF;
    a[1][1] = 0;
    for(int i = 2; i <= n; ++i){
      a[i][1] = c[1][i];
      for(int j = 2; j <= i && j <= k; ++j)
    for(int k = 1; k < i; ++k)
      if(a[i][j] > a[i-k][j-1]+c[i-k+1][i]){
        a[i][j] = a[i-k][j-1]+c[i-k+1][i];
        /*for(int i = 0; i < n; ++i) printf("%12d", i);
        printf("\n");
        for(int i = 0; i <= n; ++i){
          printf("%2d: ", i);
          for(int j = 0; j <= n; ++j) printf("%12d", a[i][j]);
          printf("\n");
        }*/
        //printf("a[%d][%d] = a[%d-%d][%d-1] + c[%d+1][%d]\n", i, j, i, k, j, k, i);
        //printf("%d = %d + %d\n", a[i][j], a[i-k][j-1], c[k+1][i]);
      }
    }
    printf("Chain %d\n", num++);
    printf("Total distance sum = %d\n\n", a[n][k]);

    scanf("%d%d", &n, &k);
  }
  return 0;
}