BZOJ 4416 阶乘字符串

Posted on 2016-09-19 17:58  ziliuziliu  阅读(84)  评论(0编辑  收藏

状压dp。状态有点难想,方程十分简单。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 455
using namespace std;
int t,n,g[maxn][23],f[(1<<22)+5],len,regis[23];
char s[maxn];
void work()
{
    memset(g,0,sizeof(g));
    memset(f,0,sizeof(f));
    scanf("%d",&n);
    scanf("%s",s);
    if (n>21) {printf("NO\n");return;}
    len=strlen(s);
    for (register int i=1;i<=21;i++) regis[i]=len+1;
    for (register int i=len-1;i>=0;i--)
    {
        for (register int j=1;j<=n;j++)
            g[i+1][j]=regis[j];
        regis[s[i]-'a'+1]=i+1;
    }
    for (register int i=1;i<=n;i++) g[0][i]=regis[i];
    for (register int i=1;i<=n;i++) g[len][i]=g[len+1][i]=len+1;
    int top=(1<<n)-1;
    for (register int i=0;i<=top;i++)
        for (register int j=1;j<=n;j++)
            if (i&(1<<(j-1)))
                f[i]=max(f[i],g[f[i^(1<<(j-1))]][j]);
    if (f[top]==len+1) printf("NO\n");
    else printf("YES\n");
}
int main()
{
    scanf("%d",&t);
    for (int i=1;i<=t;i++)
        work();
    return 0;
}