好难的dp…
好蠢的我…
勉强苟出来了,但是还不太清楚..这题很重要啊,要赶紧来补..
对于一个dp数组,每个都要遍历到,每个dp状态,如果可以列举出他所有的可以来源,并在来源处转移,可以转移到,(转移方程可以写出)没有遗漏,就是可行的.
dp[i][j],是否可以用前i个表达前j个,要不他自己是空的,要不是自己对应最后一个.
#include <iostream>
using namespace std;
#define debug(x) std::cerr << #x << " = " << (x) << std::endl
typedef long long LL;
const int MAXN = 3e2+17;
int dp[MAXN][MAXN],have[MAXN][27];
int main(int argc ,char const *argv[])
{
#ifdef noob
freopen("Input.txt","r",stdin);freopen("Output.txt","w",stdout);
#endif
int n,m,q;
cin>>n>>m>>q;
for (int i = 0; i < n; ++i)
{
string str;
cin>>str;
for (int j = 0; j < str.length(); ++j)
{
if(str[j]=='#') have[i+1][26]++;
else have[i+1][str[j]-'a']++;
}
}
while(q--)
{
string str;
cin>>str;
int l = str.length();
if(l>n)
{
cout<<"NO"<<endl;
continue;
}
for (int i = 0; i < 317; ++i)
for (int j = 0; j < 317; ++j)
dp[i][j] = 0;
dp[0][0] = 1;
for (int i = 0; i <= n-1; ++i)
{
for (int j = 0; j <= n-1; ++j)
{
if(dp[i][j])
{
if(have[j+1][26]) dp[i][j+1] = 1;
if(have[j+1][str[i]-'a']) dp[i+1][j+1] = 1;
}
}
}
if(dp[l][n]) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}