Codeforces Round #138 (Div. 2)D. Two Strings

猛的一看 觉得是dp 然后乱搞了一通 试了几种法子 要么wa了 要么tle了 后来看了别人的代码，豁然开朗。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<sstream>
#include<string>
#include<climits>
#include<stack>
#include<set>
#include<bitset>
#include<cmath>
#include<deque>
#include<map>
#include<queue>
#define iinf 0x7f7f7f7f
#define linf 1000000000000000000LL
#define dinf 1e200
#define eps 1e-11
#define all(v) (v).begin(),(v).end()
#define sz(x)  x.size()
#define pb push_back
#define mp make_pair
#define lng long long
#define sqr(a) ((a)*(a))
#define pii pair<int,int>
#define pll pair<lng,lng>
#define pss pair<string,string>
#define pdd pair<double,double>
#define X first
#define Y second
#define pi 3.14159265359
#define ff(i,xi,n) for(int i=xi;i<=(int)(n);++i)
#define ffd(i,xi,n) for(int i=xi;i>=(int)(n);--i)
#define ffl(i,r) for(int i=head[r];i!=-1;i=edge[i].next)
#define ffe(i,r) for(_edge *i=head[r];i;i=i->next)
#define cc(i,j) memset(i,j,sizeof(i))
#define two(x)          ((lng)1<<(x))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mod  1073741824
#define pmod(x,y) (x%y+y)%y
using namespace std;
typedef vector<int>  vi;
typedef vector<string>  vs;
template<class T> inline void checkmax(T &x,T y)
{
    if(x<y) x=y;
}
template<class T> inline void checkmin(T &x,T y)
{
    if(x>y) x=y;
}
template<class T> inline T Min(T x,T y)
{
    return (x>y?y:x);
}
template<class T> inline T Max(T x,T y)
{
    return (x<y?y:x);
}
template<class T> T Abs(T a)
{
    return a>0?a:(-a);
}
template<class T> inline T lowbit(T n)
{
    return (n^(n-1))&n;
}
template<class T> inline int countbit(T n)
{
    return (n==0)?0:(1+countbit(n&(n-1)));
}
char s[222222],t[222222];
int pos[26];
int len1[222222],len2[222222];
int main()
{
  while(scanf("%s%s",s+1,t+1)==2)
  {
      int n=strlen(t+1);
      int ns=strlen(s+1);
    cc(pos,-1);
    for(int i=1,j=1;i<=ns;++i)
    {if(s[i]==t[j])
    {
        pos[t[j]-'a']=j++;
    }
    len1[i]=pos[s[i]-'a'];
    }
    cc(pos,-1);
    for(int i=ns,j=n;i>=1;--i)
    {if(s[i]==t[j])
    {
        pos[t[j]-'a']=j--;
    }
    len2[i]=pos[s[i]-'a'];
    }
    bool f=1;
    ff(i,1,ns)
    if(len1[i]==-1||len2[i]==-1||len1[i]<len2[i])
    {
        f=0;break;
    }
    puts(f?"Yes":"No");
  }
    return 0;
}