USACO 4.3 Buy Low, Buy Lower(dp)

关于去重没想去怎么弄，查的题解。。。基本完全抄袭的题解。。。令dp[n+1] = 0，求dp[n+1]即可。。存4位的高精度，第一次写。。dp[i][0]存长度。

  1 /*
  2 ID:cuizhe
  3 LANG: C++
  4 TASK: buylow
  5 */
  6 #include <cstdio>
  7 #include <cmath>
  8 #include <cstring>
  9 #include <iostream>
 10 #include <map>
 11 using namespace std;
 12 int p[5003],o[5003],dp[5003][401];
 13 int next[5003];
 14 int main()
 15 {
 16     int i,n,j,k,maxz,flag = 1,z;
 17     freopen("buylow.in","r",stdin);
 18     freopen("buylow.out","w",stdout);
 19 
 20     scanf("%d",&n);
 21     for(i = 1; i <= n; i ++)
 22         scanf("%d",&p[i]);
 23     n ++;
 24     for(i = 1; i <= n; i ++)
 25     {
 26         maxz = 0;
 27         for(j = 1; j < i; j ++)
 28         {
 29             if(p[j] > p[i])
 30             {
 31                 if(maxz < o[j])
 32                     maxz = o[j];
 33             }
 34         }
 35         o[i] = maxz + 1;
 36         flag = max(o[i],flag);
 37     }
 38     for(i = 1;i < n;i ++)
 39     {
 40         for(j = i+1;j <= n;j ++)
 41         {
 42             if(p[i] == p[j])
 43             {
 44                 next[i] = j;
 45                 break;
 46             }
 47         }
 48     }
 49     for(i = 1; i <= n; i ++)
 50     {
 51         for(j = 1; j < i; j ++)
 52         {
 53             if(p[j] > p[i])
 54             {
 55                 if((next[j] == 0||next[j] > i)&&o[i] == o[j]+1)
 56                 {
 57                     int len;
 58                     len = max(dp[i][0],dp[j][0]);
 59                     for(k = 1;k <= len;k ++)
 60                     {
 61                         dp[i][k] += dp[j][k];
 62                     }
 63                     for(k = 1;k <= len;k ++)
 64                     {
 65                         if(dp[i][k] >= 10000)
 66                         {
 67                             dp[i][k+1] += dp[i][k]/10000;
 68                             dp[i][k] = dp[i][k]%10000;
 69                             if(k+1 > len)
 70                             len = k+1;
 71                         }
 72                     }
 73                     dp[i][0] = len;
 74                 }
 75             }
 76         }
 77         if(dp[i][0] == 0)
 78         {
 79             dp[i][0] = 1;
 80             dp[i][1] = 1;
 81         }
 82     }
 83     printf("%d ",flag-1);
 84     z = 1;
 85     for(i = dp[n][0];i >= 1;i --)
 86     {
 87         if(dp[n][i] >= 1000||z)
 88         printf("%d",dp[n][i]);
 89         else if(dp[n][i] >= 100)
 90         printf("0%d",dp[n][i]);
 91         else if(dp[n][i] >= 10)
 92         printf("00%d",dp[n][i]);
 93         else
 94         printf("000%d",dp[n][i]);
 95         if(i != dp[n][i])
 96         z = 0;
 97     }
 98     printf("\n");
 99     return 0;
100 }

 

posted @ 2013-03-23 19:49  Naix_x  阅读(227)  评论(0)    收藏  举报