codeforce 67C - Sequence of Balls(dp)

最新推荐文章于 2024-04-24 15:02:55 发布

原创最新推荐文章于 2024-04-24 15:02:55 发布 · 993 阅读

0 ·

CC 4.0 BY-SA版权

DP 专栏收录该内容

107 篇文章

订阅专栏

本文介绍了一个算法问题，涉及两组由小写字母标记的球序列的转换。通过四种操作：插入、删除、替换和交换相邻球，实现了从序列A到序列B的最短时间转换。文章提供了具体的限制条件及样例输入输出。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

C - Sequence of Balls

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 67C

Description

You are given a sequence of balls A by your teacher, each labeled with a lowercase Latin letter 'a'-'z'. You don't like the given sequence. You want to change it into a new sequence, B that suits you better. So, you allow yourself four operations:

You can insert any ball with any label into the sequence at any position.
You can delete (remove) any ball from any position.
You can replace any ball with any other ball.
You can exchange (swap) two adjacent balls.

Your teacher now places time constraints on each operation, meaning that an operation can only be performed in certain time. So, the first operation takes time t_i, the second one takes t_d, the third one takes t_r and the fourth one takes t_e. Also, it is given that 2· t_e ≥ t_i + t_d.

Find the minimal time to convert the sequence A to the sequence B.

Input

The first line contains four space-separated integers t_i, t_d, t_r, t_e (0 < t_i, t_d, t_r, t_e ≤ 100). The following two lines contain sequences Aand B on separate lines. The length of each line is between 1 and 4000 characters inclusive.

Output

Print a single integer representing minimum time to convert A into B.

Sample Input

Input

1 1 1 1
youshouldnot
thoushaltnot

Output

Input

2 4 10 3
ab
ba

Output

Input

1 10 20 30
a
za

Output

Hint

In the second sample, you could delete the ball labeled 'a' from the first position and then insert another 'a' at the new second position with total time 6. However exchanging the balls give total time 3.

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int mm=4e3+9;
int dp[mm][mm],pa[mm][133],pb[mm][133];
int t1,t2,t3,t4;
int la,lb;
char a[mm],b[mm];
int main()
{
  while(scanf("%d%d%d%d",&t1,&t2,&t3,&t4)!=EOF)
  { scanf("%s%s",a,b);
    la=strlen(a);lb=strlen(b);
    memset(dp,0,sizeof(dp));
    memset(pa,0,sizeof(pa));
    memset(pb,0,sizeof(pb));
    for(int i=1;i<=la;++i)
    { dp[i][0]=i*t2;///删除费用
      for(int j='a';j<='z';++j)pa[i][j]=pa[i-1][j];
      pa[i][a[i-1]]=i;///含有a[i-1]字符的最近的标号
    }
    for(int i=1;i<=lb;++i)
    {
      dp[0][i]=i*t1;///插入费用
      for(int j='a';j<='z';++j)pb[i][j]=pb[i-1][j];
      pb[i][b[i-1]]=i;
    }
    for(int i=1;i<=la;++i)
      for(int j=1;j<=lb;++j)
    { /**找出相同逆序的如a*b,b*a *为省略字符，
			  花费=交换+删除到符合条件+插入到符合条件
			  操作顺序先删，再交换，在插入
			**/
      dp[i][j]=min(dp[i-1][j]+t2,min(dp[i][j-1]+t1,dp[i-1][j-1]+t3*(a[i-1]!=b[j-1])));
      int ii=pa[i-1][b[j-1]],jj=pb[j-1][a[i-1]];
      if(ii&&jj)
      {
        dp[i][j]=min(dp[i][j],dp[ii-1][jj-1]+t4+t1*(j-1-jj)+t2*(i-1-ii));
      }
    }
    printf("%d\n",dp[la][lb]);
  }
}