题目大意:要求在同一条路上的N家快餐店,新建K个补助站点,每个快餐店和它的补助站点的距离之和最小。并且输出路径。
解题思路:这题之前想了很久,但是却漏掉最重要的一点:一条路上【1,N】快餐店,建一个补助站的话,建在中间是最优的。那么对于一个补助站是这样的,对于两个补助站的话,就看这两个补助站提供补助的范围了。dp【k】【j】表示在前j家快餐店建了k个补助站最小的补助距离。dp【k】【j】 = Min (dp【k - 1】【i】 + s【i + 1】【j】) (I>= k - 1 && i < j) 先预处理 s[i][j] ,表示i到j建个补助站最小的补助距离。
注意输出格式。restaurant(s)。
代码:
#include <cstdio> #include <cstring> #include <cstdlib> typedef long long ll; const int N = 205; const int M = 35; const ll INF = 1e13; int n, m; int d[N]; ll dis[N][N]; ll f[M][N]; int path[M][N][2]; void init () { int mid; for (int i = 1; i <= n; i++) for (int j = i; j <= n; j++) { dis[i][j] = 0; mid = (j + i) / 2; for (int k = i; k <= j; k++) dis[i][j] += labs (d[k] - d[mid]); } } void printf_ans (int k, int j) { if (!k) return; printf_ans (k - 1, path[k][j][1] - 1); if (path[k][j][1] != j) printf("Depot %d at restaurant %d serves restaurants %d to %d\n", k, path[k][j][0], path[k][j][1], j); else printf("Depot %d at restaurant %d serves restaurant %d\n", k, path[k][j][0], j); } int main () { int cas = 0; while (scanf ("%d%d", &n, &m) , n || m) { for (int i = 1; i <= n; i++) scanf ("%d", &d[i]); init (); for (int i = 1; i <= n; i++) { f[1][i] = dis[1][i]; path[1][i][0] = (1 + i) / 2; path[1][i][1] = 1; } for (int k = 2; k <= m; k++) for (int j = k; j <= n; j++) { f[k][j] = INF; for (int i = j - 1; i >= k - 1; i--) { if (f[k - 1][i] + dis[i + 1][j] < f[k][j]) { f[k][j] = f[k - 1][i] + dis[i + 1][j]; path[k][j][0] = (i + j + 1) / 2; path[k][j][1] = i + 1; } } } printf ("Chain %d\n", ++cas); printf_ans (m, n); printf ("Total distance sum = %lld\n\n", f[m][n]); } return 0; }
uva662 - Fast Food(递推),布布扣,bubuko.com
时间: 2024-10-28 16:44:12