pku1223 DEHUFF 哈夫曼树确定，PKU1261简化版

pku1223 DEHUFF 哈夫曼树确定，PKU1261简化版

题目描述：
一个哈夫曼树，给出一个字符串以及编码，要求确定哈夫曼树
如果无解或者多解，输出MULTIPLE TABLES

解法：
同pku1261，那题我写了详细的报告， http://www.cppblog.com/yzhw/archive/2011/01/11/138368.aspx

代码：

  1 # include  < cstdio >
  2 # include  < cstring >
  3 # include  < stack >
  4 using   namespace  std;
  5 struct  node
  6 {
  7    int count;
  8    node *c[2];
  9    node *pre;
 10    char end;
 11    node()
 12    {
 13        count=0;
 14        c[0]=c[1]=pre=NULL;
 15        end=0;
 16    }
 17} ;
 18 node  * head = NULL, * ans[ 27 ], * res[ 27 ];
 19 char  str[ 1024 ],code[ 1024 ];
 20 int  count,total;
 21 bool  used[ 27 ],hasfind;
 22 void  clear(node  * p = head)
 23 {
 24    if(!p) return;
 25    clear(p->c[0]);
 26    clear(p->c[1]);
 27    delete p;
 28}
 29 void  init()
 30 {
 31    clear();
 32    head=new node();
 33    head->count=1;
 34    memset(used,0,sizeof(used));
 35    memset(ans,0,sizeof(ans));
 36    count=2;
 37    hasfind=false;
 38}
 39 bool  solve( int  p1, int  p2,node  * p = head)
 40 {
 41    if(count>total||p->end) return 0;
 42    else if(str[p1]=='\0'&&code[p2]=='\0') 
 43    {
 44        if(!hasfind)
 45        {
 46            hasfind=true;
 47            memcpy(res,ans,sizeof(ans));
 48            return 0;
 49        }
 50        else return 1;
 51    }
 52    else if(str[p1]=='\0') return 0;
 53    else if(str[p1]==' '&&ans[26]||str[p1]!=' '&&ans[str[p1]-'A'])
 54    {
 55        stack<int> s;
 56        node *t=ans[str[p1]==' '?26:str[p1]-'A'];
 57        while(t->pre)
 58        {
 59            s.push(t->pre->c[0]==t?0:1);
 60            t=t->pre;
 61        }
 62        for(;code[p2]!='\0'&&!s.empty();p2++)
 63            if(code[p2]==s.top()+48)
 64                s.pop();
 65            else return 0;
 66        if(!s.empty()) return 0;
 67        else return solve(p1+1,p2);
 68    }
 69    else
 70    {
 71        p->count++;
 72        if(p->count==1)
 73        {
 74            p->end=str[p1];
 75            ans[str[p1]==' '?26:str[p1]-'A']=p;
 76            if(solve(p1+1,p2)) return 1;
 77            p->end=0;
 78            ans[str[p1]==' '?26:str[p1]-'A']=NULL;
 79        }
 80        if(code[p2]=='\0')
 81        {
 82            p->count--;
 83            return 0;
 84        }
 85        if(p->count==1)
 86            count++;
 87        if(p->c[code[p2]-'0']==NULL)
 88        {
 89            p->c[code[p2]-'0']=new node();
 90            p->c[code[p2]-'0']->pre=p;
 91        }
 92        if(solve(p1,p2+1,p->c[code[p2]-'0'])) return 1;
 93        if(p->count==1)
 94            count--;
 95        p->count--;
 96        return 0;
 97        
 98    }
 99}
100 int  main()
101 {
102    //freopen("ans.txt","w",stdout);
103    int test;
104    scanf("%d",&test);
105    getchar();
106    for(int t=1;t<=test;t++)
107    {
108        init();
109        gets(str);
110        gets(code);
111        total=0;
112        for(int i=0;str[i]!='\0';i++)
113            used[str[i]==' '?26:str[i]-'A']=true;
114        for(int i=0;i<27;i++)
115            total+=used[i];
116        printf("DATASET #%d\n",t);
117       if(solve(0,0)||!hasfind) printf("MULTIPLE TABLES\n");
118       else if(hasfind)
119       {
120           if(used[26])
121           {
122               printf("  = ");
123               stack<int> s;
124               while(res[26]->pre)
125               {
126                   s.push(res[26]==res[26]->pre->c[0]?0:1);
127                   res[26]=res[26]->pre;
128               }
129               while(!s.empty())
130               {
131                   printf("%d",s.top());
132                   s.pop();
133               }
134               printf("\n");
135           }
136           
137               for(int i=0;i<26;i++)
138                   if(used[i])
139                   {
140                       printf("%c = ",i+65);
141                       stack<int> s;
142                       while(res[i]->pre)
143                       {
144                           s.push(res[i]==res[i]->pre->c[0]?0:1);
145                           res[i]=res[i]->pre;
146                       }
147                       while(!s.empty())
148                       {
149                           printf("%d",s.top());
150                           s.pop();
151                       }
152                       printf("\n");
153                   }
154       }
155
156    }
157    return 0;
158}