POJ 1451（0ms）(dnf)（树的灵活建立）

翻译：http://blog.csdn.net/woniupengpeng/article/details/53445073

/*先建立一棵有26个节点的树，再建一个收集结果的结果集，然后用dnf去搜，与结果集相匹配的满足要求的就更新结果集，最后输出结果集就好了； */ #include <iostream>#include <fstream>#include <cstring>#include <malloc.h>using namespace std;typedef struct tree{int count;int rank;char father[100];char key[100];tree *next[26];tree(){count = 0;rank = 0;for(int i=0; i<26; ++i)next[i] = NULL;for(int i=0; i<105; ++i){father[i] = '\0';key[i] = 0;}}} tree;void dfs(tree * root);tree root;//根节点 char str[105];//输入字符串的容器 int len;//记录字符串的长度 char res[100][100]; //记录结果的单词 int res_i[100];  //比较结果的概率 char str_arr[] = "22233344455566677778889999";int main(){int row,num,flag,max,sum,now;int input;tree *temp;scanf("%d",&sum);for(int sum1=1; sum1<=sum; ++sum1){printf("Scenario #%d:\n",sum1);//初始化根节点 root.count = 0;for(int i=0; i<26; ++i){root.next[i] = NULL;}//初始化根节点END//接收输入scanf("%d",&row);while(row--){temp = &root;scanf("%s",str);scanf("%d",&num);len = strlen(str);for(int i=0; i<len; ++i){int index = str[i] - 97;if(temp->next[index] == NULL){tree * t1 = new tree();temp->next[index] = t1;}//更新结构体里的各种信息； temp->next[index]->count += num;temp->next[index]->rank = i+1;strcpy(temp->next[index]->father, str);temp->next[index]->father[i+1] = '\0';strcpy(temp->next[index]->key, temp->key);temp->next[index]->key[i] = str_arr[str[i]-97];temp->next[index]->key[i+1] = '\0';//让更新数据的节点，指向下一个节点； temp = temp->next[index];}}//测试阶段scanf("%d",&row);for(int i=0; i<row; ++i){scanf("%s",str);len = strlen(str);for(int j=0; j<len; ++j){res_i[j] = 0;for(int k=0; k<len; ++k){res[j][k] = '\0';}}str[len-1] = '\0';dfs( &root);for(int i=0; i<len-1; ++i){if(strlen(res[i]) !=0 ){printf("%s\n",res[i]);}elseprintf("MANUALLY\n");}printf("\n");}printf("\n");}}void dfs(tree *node){char str1[100];int i;for(i=0; i<node->rank+1; ++i){str1[i] = str[i];}str1[i] = '\0';for(i=0; i<26; ++i){if(node->next[i] != NULL){//只有九宫格数字匹配和单词的概率大于结果集里的就可以更新； if( (res_i[node->next[i]->rank-1] < node->next[i]->count) && strcmp(str1, node->next[i]->key)==0 ){res_i[node->next[i]->rank-1] = node->next[i]->count;strcpy(res[node->next[i]->rank-1],node->next[i]->father);}if(node->next[i]->rank+1 < len)dfs(node->next[i]);}}}