pat 1015. Letter-moving Game (35)(LCS)

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1015. Letter-moving Game (35)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CAO, Peng

Here is a simple intersting letter-moving game. The game starts with 2 strings S and T consist of lower case English letters. S and T contain the same letters but the orders might be different. In other words S can be obtained by shuffling letters in String T. At each step, you can move one arbitrary letter in S either to the beginning or to the end of it. How many steps at least to change S into T?

Input Specification:

Each input file contains one test case. For each case, the first line contains the string S, and the second line contains the string T. They consist of only the lower case English letters and S can be obtained by shuffling T's letters. The length of S is no larger than 1000.

Output Specification:

For each case, print in a line the least number of steps to change S into T in the game.

Sample Input:

iononmrogdg
goodmorning

Sample Output:

8

Sample Solution:

(0) starts from iononmrogdg
(1) Move the last g to the beginning: giononmrogd
(2) Move m to the end: giononrogdm
(3) Move the first o to the end: ginonrogdmo
(4) Move r to the end: ginonogdmor
(5) Move the first n to the end: gionogdmorn
(6) Move i to the end: gonogdmorni
(7) Move the first n to the end: googdmornin
(8) Move the second g to the end: goodmorning

因为可以将字母放到最前面和最后面 ，那么LCS就要保证模式串匹配的是文本串的一段连续的字串

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1010;
int dp[N][N], a[N];
char s1[N], s2[N];

int main()
{
    memset(dp,0,sizeof(dp));
    scanf("%s %s", s2+1, s1+1);

    int l1=strlen(s1+1), l2=strlen(s2+1);
    for(int i=1;i<=l1;i++)a[i]=l1+1;
    int ans=0;
    for(int i=1;i<=l1;i++)
    {
        for(int j=1;j<=l2;j++)
        {
            if(s1[i]==s2[j]&&j>a[i-1])
                dp[i][j]=dp[i-1][j-1]+1,a[i]=min(a[i],j);
            else if(s1[i]==s2[j])
                dp[i][j]=1,a[i]=min(a[i],j);
            else dp[i][j]=dp[i][j-1];
            ans=max(ans,dp[i][j]);
        }
    }
    printf("%d\n",l1-ans);
    return 0;
}