CodeForces 157C Message

$dp$。

$dp[i][j]$表示$s[i]$到$s[j]$和$t[lent-1+i-j]$到$t[lent-1]$有$dp[i][j]$位相同，然后枚举一遍$dp[i][j]$就可以算出答案了。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;

const int maxn=5100;
char s[maxn],t[maxn];
int lens,lent;
int dp[maxn][maxn];

int main()
{
    memset(s,0,sizeof s);
    memset(t,0,sizeof t);
    scanf("%s%s",s,t);
    lens=strlen(s), lent=strlen(t);

    for(int i=lens;i<lens+lent;i++) s[i]='.';
    lens=strlen(s);
    
    int ans=5000;

    for(int Len=1;Len<=lent;Len++)
    {
        for(int i=0;i<lens;i++)
        {
            int j=i+Len-1;
            if(j>=lens) continue;
            if(j==i)
            {
                if(s[i]==t[lent-1]) dp[i][j]=1;
                else dp[i][j]=0;
                ans=min(ans,j-i+1-dp[i][j]+lent-(j-i+1));
            }
            else
            {
                dp[i][j]=dp[i+1][j];
                if(s[i]==t[lent-1+i-j]) dp[i][j]++;
                ans=min(ans,j-i+1-dp[i][j]+lent-(j-i+1));
            }
        }
    }

    printf("%d\n",ans);

    return 0;
}