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)%yusing 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;}