BZOJ5073 小A的咒语（动态规划）

　　设f[i][j][0/1]为前i位选j段时其中第i位选/不选最多能匹配到哪，转移时f[i][j][0]→f[i+1][j][0]，f[i][j][1]→f[i+1][j][0]，f[i][j][1]→f[i+1][j][1]，f[i][j][0]→f[i+1][j+1][1]。失配时找到最后一位相同字符，具体见代码。感觉非常假，欢迎hack。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
#define M 102
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int T,n,m,t,f[N][M][2],pre[N][26];
char a[N],b[N];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5073.in","r",stdin);
    freopen("bzoj5073.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read();
    while (T--)
    {
        n=read(),m=read(),t=read();
        scanf("%s",a+1),scanf("%s",b+1);
        for (int i=0;i<26;i++) pre[0][i]=-1;
        for (int i=1;i<=m;i++)
        {
            for (int j=0;j<26;j++)
            pre[i][j]=pre[i-1][j];
            pre[i][b[i]-'a']=i;
        }
        f[0][0][1]=-1;
        bool flag=0;
        for (int i=1;i<=n;i++)
        {
            f[i][0][1]=-1;
            for (int j=1;j<=t;j++)
            {
                f[i][j][0]=max(f[i-1][j][0],f[i-1][j][1]);
                f[i][j][1]=pre[max(f[i-1][j-1][0],f[i-1][j][1])+1][a[i]-'a'];
                if (f[i][j][1]==m) {flag=1;break;}
            }
            if (flag) break;
        }
        if (flag) puts("YES");else puts("NO");
    }
    return 0;
}