uva 662 Fast Food （dp）

本文出自   http://blog.youkuaiyun.com/shuangde800

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题意

一条直线马路上有n个饭店，各个坐标为di.

要在n个饭店中选择k个饭店用来建造停车场。没有建停车场的饭店，只能使用附近最近的一个停车场。

问总距离最少的建造方案，并输出。

思路

先进行预处理，sum[i][j]表示在饭店i～j之间建一个停车场，i～j的所有饭店到停车场的距离之和最小。

在饭店i～j之间，选择在(i+j)/2点建造是总距离最小的方案

f[i][j]，表示前i个饭店，建造j个停车场的最小总距离
那么，
f[i][j] = min{ f[k-1][j] + sum[k][i], 1<=k<=i }

至于输出方案，dp的输出方案一般都是保存“决策”记录，然后递归输出即可。

代码

/**==========================================
 *   This is a solution for ACM/ICPC problem
 *
 *   @source：uva-662 Fast Food
 *   @type:  dp
 *   @author: shuangde
 *   @blog: blog.youkuaiyun.com/shuangde800
 *   @email: zengshuangde@gmail.com
 *===========================================*/
#include
   
   
    
    
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           using namespace std; typedef long long int64; const int INF = 0x3f3f3f3f; const int MAXN = 210; int n, k; int d[MAXN], f[MAXN][35], sum[MAXN][MAXN]; int mark[MAXN][MAXN], mid[MAXN][MAXN]; int idx; void print(int i, int j) { if (i < 1 || j < 1) return; print(mark[i][j] - 1, j - 1); printf("Depot %d at restaurant %d serves restaurant", idx++, mid[mark[i][j]][i]); if (mark[i][j] == i) { printf(" %d\n", i); } else { printf("s %d to %d\n", mark[i][j], i); } } int main() { int cas = 1; while (~scanf("%d%d", &n, &k) && n + k) { for (int i = 1; i <= n; ++i) scanf("%d", &d[i]); memset(sum, 0, sizeof (sum)); for (int i = 1; i <= n; ++i) { mid[i][i] = i; for (int j = i + 1; j <= n; ++j) { int m = (j + i) >> 1; mid[i][j] = m; for (int k = i; k <= j; ++k) { sum[i][j] += abs(d[k] - d[m]); } } } for (int i = 1; i <= n; ++i) for (int j = 0; j <= k; ++j) f[i][j] = INF; memset(f[0], 0, sizeof (f[0])); for (int i = 1; i <= n; ++i) for (int j = 1; j <= k; ++j) { for (int k = 1; k <= i; ++k) { int tmp = f[k - 1][j - 1] + sum[k][i]; if (tmp <= f[i][j]) { mark[i][j] = k; f[i][j] = tmp; } } } printf("Chain %d\n", cas++); idx = 1; print(n, k); printf("Total distance sum = %d\n\n", f[n][k]); } return 0; }