• Topcoder SRM 698 Div1 250 RepeatString(dp)


    题意

    [题目链接]这怎么发链接啊。。。。。

    Sol

    枚举一个断点,然后类似于LIS一样dp一波

    这个边界条件有点迷啊。。fst了两遍。。。

    
    #include<bits/stdc++.h>
    using namespace std;
    const int MAXN = 1e5 + 10, INF = 1e9 + 7;
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N, f[101][101];
    class RepeatString{
    public:
        int solve(int pos, string s) {
            string a, b; a += '.'; b += '*';
            memset(f, 0x3f, sizeof(f));
            for(int i = 0; i < pos; i++) a += s[i];
            for(int i = pos; i < N; i++)  b += s[i];
           
            int la = a.length() - 1, lb = b.length() - 1;
            f[0][0] = 0;
            for(int i = 0; i <= la; i++) f[i][0] = i;
            for(int i = 0; i <= lb; i++) f[0][i] = i;
            for(int i = 1; i <= la; i++) {
                for(int j = 1; j <= lb; j++) {
                    f[i][j] = min(f[i - 1][j] + 1, f[i][j - 1] + 1);
                    f[i][j] = min(f[i][j], f[i - 1][j - 1] + (a[i] != b[j]));
                }
            }
            return f[la][lb];
        }
        int minimalModify(string s) {
            N = s.length(); int ans = N;
            for(int i = 0; i < N; i++) ans = min(ans, solve(i, s));
            return ans;
        }
    };
    
    int main() {
        string s; 
        cin >> s;
        cout << RepeatString().minimalModify(s);
    }
    /*
    pkafkbeabccfjejkdgkaatcedaocgmecaapakfvbfgefr
    */
    
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/9775254.html
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