HDOJ - 4545 魔法串 简单DP

     用二维bool记录可行的对应关系..为了处理方便..预处理所有的f[i][i]=true....

    dp[i]代表以b[i]为最后一个字符能匹配到a串的最大长度...

Program:

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
#define ll long long
#define oo 1000007
#define MAXN 1005
using namespace std; 
char a[MAXN],b[MAXN];
int dp[MAXN];
bool f[26][26];
int main()
{    
       freopen("input.txt","r",stdin);   freopen("output.txt","w",stdout);  
       int T,t,i,j,m,la,lb;
       bool ans;
       char c1,c2;
       scanf("%d",&T);
       for (t=1;t<=T;t++)
       {
              scanf("%s%s%d",a+1,b+1,&m);
              memset(f,false,sizeof(f));
              for (i=0;i<26;i++) f[i][i]=true;
              while (m--)
              {
                     do { c1=getchar(); } while (c1<'a' || c1>'z');
                     do { c2=getchar(); } while (c2<'a' || c2>'z');  
                     f[c1-'a'][c2-'a']=true;
              }
              memset(dp,0,sizeof(dp));
              la=strlen(a+1);     lb=strlen(b+1);
              ans=false;
              for (i=1;i<=lb;i++)
                 for (j=0;j<i;j++)
                    if (dp[i]<dp[j]+1 && f[b[i]-'a'][a[dp[j]+1]-'a'])
                    {
                           dp[i]=dp[j]+1;
                           if (dp[i]==la)
                           {
                                 ans=true;
                                 goto A;
                           }
                    }
              A: ;
              printf("Case #%d: ",t);
              if (ans) printf("happy\n");
                 else  printf("unhappy\n");
       }
       return 0;
}