DP(递归打印路径) UVA 662 Fast Food

最新推荐文章于 2022-11-22 18:22:51 发布

weixin_34320159

最新推荐文章于 2022-11-22 18:22:51 发布

阅读量97

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Running-Time/p/4773925.html

本文提供了一种解决 UVA 662 题目的算法思路，利用动态规划方法寻找最优的停车场位置，使得所有饭店到停车场的最短距离之和最小。文章还介绍了递归打印路径的方法。

题目传送门

题意：n个饭店在一条直线上，给了它们的坐标，现在要建造m个停车场，饭店没有停车场的要到最近的停车场，问所有饭店到停车场的最短距离

分析：易得区间(i, j)的最短距离和一定是建在(i + j) / 2的饭店，预处理出(i, j)的距离和sum[i][j]，mark[i][j] 表示区间的最优停车场的位置，mid[i][j]表示(i + j) / 2。状态转移方程：dp[i][j] = max (dp[k-1][j-1] + sum[k][i])；

收获：学习递归打印路径

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-29 16:42:33
* File Name     :UVA_662.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 2e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int dp[N][33];
int sum[N][N], d[N], mark[N][N], mid[N][N];
int n, m;
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); return ;
    }
    else    printf ("s %d to %d\n", mark[i][j], i);
}

int main(void)    {
    int cas = 0;
    while (scanf ("%d%d", &n, &m) == 2) {
        if (!n && !m)   break;
        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 mm = (i + j) >> 1;
                mid[i][j] = mm;
                for (int k=i; k<=j; ++k)    {
                    sum[i][j] += abs (d[mm] - d[k]);
                }
            }
        }
        memset (dp, INF, sizeof (dp));
        memset (dp[0], 0, sizeof (dp[0]));
        for (int i=1; i<=n; ++i)    {
            for (int j=1; j<=m; ++j)    {
                for (int k=1; k<=i; ++k)    {
                    int tmp = dp[k-1][j-1] + sum[k][i];
                    if (dp[i][j] >= tmp) {
                        dp[i][j] = tmp;
                        mark[i][j] = k; 
                    }
                }
            }
        }
        printf ("Chain %d\n", ++cas);
        idx = 0;    print (n, m);
        printf ("Total distance sum = %d\n\n", dp[n][m]);
    }

    return 0;
}

转载于:https://www.cnblogs.com/Running-Time/p/4773925.html