uva 297

//这题用的是最笨的方法，建树然后合并，然后计算//最开始建树都有点问题，最开始用的是指针指向字符串，通过指针一个一个建树，可是怎么都过不了，于是就参考网上直接传数组，求大神解释一下啊//还有一个问题就是 最后数据都过了总是runtime error，果然一般runtime error都是数组开小了,,,把数组从1000开到5000  终于AC!!!!!!!!#include <stdio.h>#include <malloc.h>#include <string.h>typedef struct tree{  char val;  struct tree* son[4];}*Tree, T;int sum, pixel, loc;void dfs( Tree root, int pixel ){  int i;  if( root->val == 'f' )    sum += pixel;  for( i = 0; i < 4; i++ )  if( root->son[i] )    {      pixel /= 4;      dfs( root->son[i], pixel );      pixel *= 4;    }  return ;}Tree newnode(char a[],int terminal){int i;    Tree root;    root=(Tree)malloc(sizeof(T));    if(loc < terminal )    {        root->val=a[loc];         if(a[loc]=='p')        {for( i = 0; i < 4; i++ ){++loc;root->son[i] = newnode( a, terminal );}        }        else for( i = 0; i < 4; i++ ){root->son[i] = NULL;}    }    return root;}Tree combine(Tree root1,Tree root2){int i;    Tree root=(Tree)malloc(sizeof(T));    if(root1->val=='f'||root2->val=='f')    {        root->val='f';        for( i = 0; i < 4; i++ )root->son[i] = NULL;    }    else if(root1->val=='e'&&root2->val=='e')    {        root->val='e';        for( i = 0; i < 4; i++ )root->son[i] = NULL;    }    else    {        root->val='p';        if(root1->val=='p'&&root2->val=='p')        {for( i = 0; i < 4; i++ )            root->son[i]=combine(root1->son[i],root2->son[i]);        }        else if(root1->val='p'&&root2->val=='e')        {            for( i = 0; i < 4; i++ )root->son[i] = root1->son[i];        }        else         {            for( i = 0; i < 4; i++ )root->son[i] = root2->son[i];          }    }    return root;}int main(){  char s1[5000];  char s2[5000];  int len1, len2;  Tree root1, root2, root;  int N;  scanf( "%d", &N );  getchar();  while( N-- )    {      loc = 0;      sum = 0;      pixel = 1024;      scanf( "%s", s1 );      scanf( "%s", s2 );      root1 = newnode( s1, strlen(s1) );      loc = 0;      root2 = newnode( s2, strlen(s2) );      root = combine( root1,  root2);      dfs( root, pixel );     printf( "There are %d black pixels.\n", sum );    }  return 0;}