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  • 动态规划----最长公共子序列

    给定连个字符串s1s2... ...sn和t1t2... ...tn。求出这两个字符串最长的公共子序列的长度。字符串s1s2... ...sn的子序列可以表示为si1si2... ...sim(i1<i2<... ...<sim)的序列。(1<=n,m<=1000)

    dp[i][j]表示s1... ... si和t1... ...tj对应的LCS的长度。
    当s[i+1]==t[i+1]时,dp[i+1][j+1] = max(dp[i][j]+1, dp[i+1][j], dp[i][j+1]);//其实只要dp[i+1][j+1]=dp[i][j]+1 就可以了。
    其它情况,dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1]);
     

    #include<cstdio>
    using namespace std;
    const int MAXN=1001;
    int char s[MAXN],t[MAXN];
    int n,m;
    int solve()
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
            {
                if(s[i-1]==t[i-1])
                    dp[i][j]=dp[i-1][j-1]+1;
                else
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
            }
        return dp[n][m];
    }
    int main()
    {
        scanf("%d%d",&n,&m);
        scanf("%s%s",s,t);
        printf("%d",solve);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ke-yi-/p/10175829.html
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