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DP(递归打印路径) UVA 662 Fast Food

2015-08-31 19:41 375 查看
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
[33];
int sum

, d
, mark

, mid

;
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
[m]);
}

return 0;
}


  
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