题目如下：

#include <iostream>

#include <string>

#include <vector>

using namespace std;

// use this struct to store square subsequence, 4 positions and 1 length

struct SqSb {

    // take square subsequence as two subsquence s0 and s1

    int s00; // the position of s0's first char

    int s01; // the position of s0's last char

    int s10;

    int s11;

    int len;

    SqSb() {

        s00 = s01 = s10 = s11 = 0;

        len = 0;

    }

    SqSb(int t00, int t01, int t10, int t11, int length) {

        s00 = t00;

        s01 = t01;

        s10 = t10;

        s11 = t11;

        len = length;

    }

};

int maxSqSubLen(const string & str) {

    int strLen = str.size();

    // corner cases

    if (strLen < 1) return 0;

    if (strLen == 2) {

        if (str[0] == str[1]) return 2;

        else return 0;

    }

    // corner cases end

    // dp[i] stores the square subsequence of length (i + 1) * 2

    vector<vector<SqSb> > dp;

    // dp1 == dp[0] is the initial data

    vector<SqSb> dp1;

    for (int i = 0; i < strLen - 1; ++i) {

        char ich = str[i];

        for (int j = i + 1; j < strLen; ++j) {

            if (ich == str[j]) {

                SqSb s(i, i, j, j, 2);

                dp1.push_back(s);

            }

        }

    }

    // there is no duplicate char in this string return

    if (dp1.empty()) return 0;

    dp.push_back(dp1);

    for (int l = 2; l <= strLen/2; ++l) {

        vector<SqSb> dpl;

        for (int i = 0; i < dp[l - 2].size(); ++i) {

            SqSb si = dp[l - 2][i];

            for (int j = 0; j < dp1.size(); ++j) {

                SqSb sj = dp1[j];

                if (sj.s00 > si.s01 && sj.s00 < si.s10

                && sj.s10 > si.s11) {

                    SqSb s(si.s00, sj.s00, si.s10, sj.s10, l * 2);

                    dpl.push_back(s);

                }

            }

        }

        if (dpl.empty()) return (l - 1) * 2;

        dp.push_back(dpl);

    }

    return strLen/2 * 2;

}

int main(int argc, char **argv) {

    cout << maxSqSubLen(string(argv[1])) << endl;

    return 0;

}

参考的是 * 的一个提问：https://*.com/questions/10000226/square-subsequence

题目不难，知道DP的整体流程，但是分析问题的能力差了一点。