HDU 4362 Dragon Ball (DP)

 

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=4362

 

设dp[i][j]表示第i批龙珠中取第j个需要花费的最小体力。

dp[i][j] = min{ dp[i-1][k] + abs(pos[i-1][k]-pos[i][j]) } + cost[i][j];

官方题解说这样会超时，但是是可以蹭过去的。。 

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4  const  int N= 51;
 5  const  int M= 1001;
 6  const  int inf= 0x7fffffff;
 7  int pos[N][M],cost[N][M],dp[N][M];
 8  int main()
 9 {
10      int t,n,m,i,j,k,x,min,ab;
11     scanf( " %d ",&t);
12      while(t--)
13     {
14         scanf( " %d%d%d ",&n,&m,&x);
15          for(i= 1;i<=n;i++)
16              for(j= 1;j<=m;j++)
17                 scanf( " %d ",&pos[i][j]);
18          for(i= 1;i<=n;i++)
19              for(j= 1;j<=m;j++)
20                 scanf( " %d ",&cost[i][j]);
21          for(i= 1;i<=m;i++)
22         {
23             ab=x-pos[ 1][i];
24              if(ab< 0)
25                 ab=-ab;
26             dp[ 1][i]=ab+cost[ 1][i];
27         }
28          for(i= 2;i<=n;i++)
29         {
30              for(j= 1;j<=m;j++)
31             {
32                 min=inf;
33                  for(k= 1;k<=m;k++)
34                 {
35                     ab=pos[i][j]-pos[i- 1][k];
36                      if(ab< 0)
37                         ab=-ab;
38                      if(min>dp[i- 1][k]+ab+cost[i][j])
39                         min=dp[i- 1][k]+ab+cost[i][j];
40                 }
41                 dp[i][j]=min;
42             }
43         }
44         min=inf;
45          for(i= 1;i<=m;i++)
46         {
47              if(min>dp[n][i])
48             min=dp[n][i];
49         }
50         printf( " %d\n ",min);
51     }
52      return  0;
53 }