Codeforces1303E. Erase Subsequences

转换一下题意，就相当于问t能不能和s中2个不相同的子串相同,我们可以将t串拆成2个子串t1,t2，得到状态dp[i][j][k]=0/1，表示s判断到i位，t1判断到j位，t2判断到k位，0/1表示是否满足

两个状态,s[i]与t1[j]相匹配，s[i]与t2[k]相匹配

dp[i+1][j+1][k] = dp[i][j][k]

dp[i+1][j][k+1] = dp[i][j][k]

因为dp的值只取0和1，我们可以优化dp函数，dp[i][j][k]可以变成dp[i][j] = k，这样可以消除一维，最后判断dp[s.size()][t1.size()]是否>=t2.size()，也就是匹配成功即可，复杂度为O(n^3)

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

int dp[405][405];

bool check(string s, string t) {
    int idx = 0;
    if(!t.size()) return true;
    for(int i = 0; i < s.size(); ++i) {
        if(s[i] == t[idx]) idx++;
        if(idx == t.size()) return true;
    }
    return false;
}

bool check(string s, string t1, string t2) {
    memset(dp, -1, sizeof(dp));
    dp[0][0] = 0;
    for(int i = 0; i < s.size(); ++i)
        for(int j = 0; j <= t1.size(); ++j)
            if(dp[i][j] >= 0) {
                if(j<t1.size()&&s[i] == t1[j]) dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j]);
                if(dp[i][j]<t2.size()&&s[i] == t2[dp[i][j]]) dp[i+1][j] = max(dp[i+1][j], dp[i][j]+1);
                dp[i+1][j] = max(dp[i][j], dp[i+1][j]);
            }
    int s1 = dp[s.size()][t1.size()], s2 = t2.size();
    if(s1 >= s2) return true;
    return false;
}

void run_case() {
    string s, t;
    cin >> s >> t;
    /* if(check(s, t)) {
        cout << "YES
";
        return;
    } */
    for(int i = 0; i < t.size(); ++i) {
        string t1 = "", t2 = "";
        for(int j = 0; j <= i; ++j) {
            t1 += t[j];
        }
        for(int j = i+1; j < t.size(); ++j)
            t2 += t[j];
        if(check(s, t1, t2)) {
            cout << "YES
";
            return;
        }
    }
    cout << "NO
";
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    //cout.setf(ios_base::showpoint);cout.precision(10);
    int t; cin >> t;
    while(t--)
    run_case();
    cout.flush();
    return 0;
}