Approximate Search

题目链接：Gym-101492H

动态规划，应该是比较基础的，可是自己就是不会QAQ....

 /* 把使用机会当成“花费” */
 # include <iostream>
 # include <cstdio>
 # include <cstring>
 # include <string>
 # include <cstdlib>
 # include <cmath>
 # include <ctime>
 # include <climits>
 # include <memory>
 # include <functional>
 # include <algorithm>
 # include <bitset>
 # include <set>
 # include <map>
 # include <stack>
 # include <vector>
 # include <deque>
 # include <queue>
 # include <iomanip>
 # include <utility>
 using namespace std;

 # define lson l,m,rt<<
 # define rson r,m+,rt<<|
 # define lowbit(x) (x&(-x))
 # define lcm(a,b) (a*b/__gcd(a,b))
 typedef long long ll;
 const ll mod=1e9+;
 const int maxn=1e6+;
 const double pi=acos(-1.0);
 const int eps=1e-;

 char s[], t[];
 int dp[maxn][];
 //dp[i][j]表示t[1-i]和t[1-j]能比配需要的最小的k值
 int main()
 {
     int n, m, k;
     memset(dp, , sizeof(dp));
     cin>>m>>n>>k;
     scanf("%s", s+);
     scanf("%s", t+);

     //如果模式串s是空串,则k为0
     for(int i=; i<=n; i++ )
         dp[i][] = ;

     for(int i=; i<=n; i++ )//t
     {
         for(int j=; j<=m; j++ )//s
         {
             //s中前(j-1)个和t中前(i-1)个匹配后,s[j]和t[i]直接匹配
             if( s[j]==t[i] )
                 dp[i][j] = min(dp[i-][j-], dp[i][j]);//不需要”花费“

                 //在前面已经匹配的基础上,删掉t[i],此时s[j]依旧没有匹配上,所以是dp[i][j-1],而不是dp[i][j]
                 dp[i][j-] = min(dp[i][j-], dp[i-][j-]+);
                 //在前面已经匹配的基础上,增加一个t[i]与s[j]匹配,但由于是虚拟的增加,所以仍是dp[i-1][j],而不是dp[i][j];
                 dp[i-][j] = min(dp[i-][j], dp[i-][j-]+);
                 //在前已经匹配的基础上,换t[i]
                 dp[i][j] = min(dp[i][j], dp[i-][j-]+);
         }
     }

     for(int i=; i<=n; i++ )
     {
         if( dp[i][m]<=k )
         {
             cout<<'S'<<endl;
             return ;
         }
     }
     cout<<'N'<<endl;
     return ;
 }