662 - Fast Food

做这道题时没有什么想法，后来在《使用算法分析与程序设计》上看到了这道题，除了输出完全一样。看了以后，觉得自己好好想想也应该能做出来的，思维还是不够开阔。
设w[i][j]表示从i到j只放一个depots的最小距离，w[i][j] = ∑|d[l] - d[(i + j) / 2]| (i <= l <= j)。设状态m[i][j]表示前j个饭店放i个depots的最小距离和，则m[k]
即为问题的解。
边界条件：m[1][j] = w[1][j] (1 <= j <= n)；状态转移方程：m[i][j] = min(m[i - 1][l] + w[l + 1][j]) (i - 1 <= l <= j - 1)，时间复杂度为O(k*n2)。
最后递归输出即可，做的时候输出上浪费了很多时间。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int INF = (1<<29);
int d[205], w[205][205], m[35][205], v[205][205];
void print_ans(int i, int j, int sum, int cur)
{
if(cur == 0) return ;
for(int loop = i - 1; loop < j; loop++){
if(m[i - 1][loop] + w[loop + 1][j] == sum){
print_ans(i - 1, loop, sum - w[loop + 1][j], cur - 1);
if(loop + 1 == j)
printf("Depot %d at restaurant %d serves restaurant %d\n", cur, (j + loop + 1) / 2, j);
else printf("Depot %d at restaurant %d serves restaurants %d to %d\n", cur, (j + loop + 1) / 2, loop + 1, j);
break;
}
}
}

int main()
{
//freopen("input.txt", "r", stdin);
int n, k;
int con = 1;
while(scanf("%d %d", &n, &k) == 2){
if(!n && !k) break;
for(int i = 1; i <= n; i++)  scanf("%d", &d[i]);
memset(w, 0, sizeof(w));
for(int i = 1; i <= n; i++){
for(int j = i + 1; j <= n; j++){
int sum = 0;
for(int loop = i; loop <= j; loop++)
sum += (int)fabs((d[loop] - d[(j + i) / 2]));
w[i][j] = w[j][i] = sum;
}
}
for(int j = 1; j <= n; j++) m[1][j] = w[1][j];
for(int i = 2; i <= k; i++){
for(int j = i; j <= n; j++){
m[i][j] = INF;
int x;
for(int loop = i - 1; loop <= j - 1; loop++){
m[i][j] = min(m[i][j], m[i - 1][loop] + w[loop + 1][j]);
}
}
}
printf("Chain %d\n", con++);
print_ans(k, n, m[k]
, k);
printf("Total distance sum = %d\n\n", m[k]
);
}
return 0;
}