uva 297

来源：互联网发布：紫阳真人知乎编辑：程序博客网时间：2024/06/11 10:50

题意：出四叉树的前序遍历，计算2个树叠加后黑色的面积大小，每个节点要是为叶子节点要么必有四个子节点。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;char str1[100005],str2[100005];class  Node{public:char data;Node *son[4];};Node node[20000];int nIndex,pos1,pos2,sum;inline Node* NewNode(){node[nIndex].data = 0 ;for (int i = 0 ; i < 4 ; i++)node[nIndex].son[i] = NULL;return &node[nIndex++];}Node *BuildTree(Node *root,int &pos,char *str){pos++;if (pos == strlen(str))return NULL;root = NewNode();root->data = str[pos];if (str[pos] == 'p'){for(int i = 0 ; i < 4 ; i++ )if (root->son[i] == NULL)root->son[i] = BuildTree(root->son[i],pos,str);}return root;}void dfs(Node *root1,Node *root2,int level){if (root1 == NULL && root2 == NULL )return ;if ( root1 == NULL ){if(root2->data == 'f'){sum+= 1024>>(level*2);return ;}for (int i = 0; i < 4 ; i++ )dfs(root1,root2->son[i],level+1);return ;}if (root2 == NULL ){if (root1->data == 'f'){sum+= 1024>>(level*2);return ;}for (int j = 0 ; j < 4 ; j++ )dfs(root1->son[j],root2,level+1);return ;}if (root1->data=='f' || root2->data == 'f'){sum+=1024>>(level*2);return ;}    for (int l = 0 ; l < 4 ; l++ )dfs(root1->son[l],root2->son[l],level+1);}void solve(){Node *root1 = NULL,*root2 = NULL;nIndex = 0;pos1 = pos2 = -1;root1 = BuildTree(root1,pos1,str1);root2 = BuildTree(root2,pos2,str2);dfs(root1,root2,0);printf("There are %d black pixels.\n",sum);}int main(){int T;cin>>T;getchar();while (T--){scanf("%s %s",str1,str2);sum = 0 ;solve();}return 0 ;}