LightOJ 1051 Good or Bad

题意：给你一个字符串，只包含大写字母和‘？’，如果字符串中出现了连续三个以上的元音字母或者连续五个以上的辅音字母，则这个字符串是bad,不然就是good.

‘？’号可以替换任意字母，即可bad有可good，则输出Mixed.

分析：dpy[i][j]表示到i为止，连续j个元音可否达到。dpf表示辅音。

            这个转移当达到3个元音或者5个辅音时后面就不在转移了，即全是零。  所以找是否是mixed只需扫最后一个位置的前两个元音或者前4个辅音是否不为0。

            不懂可以打印下屏蔽的代码，研究下。

#include<iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<cctype>#include<set>#include<map>#include<queue>#include<stack>#include<iomanip>#include<sstream>#include<limits>#define ll long long#define inf 0x3f3f3f3fusing namespace std;const int N= 2000;const int maxn = 1e2+10;const int maxm = 2000000001;const int MOD = 1000;int dpy[maxn][maxn],dpf[maxn][maxn];   //元音  辅音  到i为止，连续j个元音可否达到char str[maxn];int main(){#ifdef LOCALfreopen("C:\\Users\\lanjiaming\\Desktop\\acm\\in.txt","r",stdin);//freopen("output.txt","w",stdout);#endifios_base::sync_with_stdio(0);    int T,kase = 0;    cin>>T;    while(T--)    {        cout<<"Case "<<++kase<<": ";        cin>>str;        int len = strlen(str);        memset(dpy,0,sizeof(dpy));        memset(dpf,0,sizeof(dpf));        for(int i = 0; i < len; i++)        {            int x;            if (str[i] == '?') x = 0;            else if (str[i] == 'A' || str[i]=='E' || str[i]=='I' || str[i]=='O'||str[i]=='U')                  x = 1;                 else x = 2;            if (i==0)            {                if ( x == 1 || x == 0) dpy[0][1] = 1;                if ( x == 2 || x == 0) dpf[0][1] = 1;            }else             {                 for(int j = 0; j <= 2; j++)                    if (dpy[i-1][j])                    {                         if ( x == 1 || x == 0) dpy[i][j+1] = 1;                         if ( x == 2 || x == 0) dpf[i][1] = 1;                    }                 for(int j = 0; j <= 4; j++)                    if (dpf[i-1][j])                    {                         if ( x == 1 || x == 0) dpy[i][1] = 1;                         if ( x == 2 || x == 0) dpf[i][j+1] = 1;                    }             }        }       /* for(int i = 0; i < len; i++)        {            cout<<i<<endl;            for(int j = 0; j <= 3; j++) cout<<dpy[i][j]<<" ";cout<<endl;            for(int j = 0; j <= 5; j++) cout<<dpf[i][j]<<" ";cout<<endl;        }*/        bool ok1 =false,ok2 = false;        for(int i = 0; i < len; i++)           if (dpy[i][3] || dpf[i][5]) ok2 =true;                   for(int i = 0; i <= 2; i++)             if (dpy[len-1][i]) ok1 =true;        for(int i = 0; i <= 4; i++)             if (dpf[len-1][i]) ok1 =true;                     if (ok1 && ok2 ) cout<<"MIXED\n";        else if (ok1) cout<<"GOOD\n";              else cout<<"BAD\n";    }return 0;}