POJ_2418_Hardwood Species(map实现、Trie tree实现)

来源：互联网发布：cmseasy 知乎编辑：程序博客网时间：2024/06/09 21:46

题型：综合题，各种方法解决。

分析：

感觉这是一个强大的题目，可以用各种高难度数据结构解决。

思想很简单，就不多说了。

代码：

map实现：

#include <iostream>#include <string>#include <map>#include <iterator>#include <cstdio>using namespace std;int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    string s;    int cnt = 0;    map<string,int> tree;    while(getline(cin,s)!=NULL){    tree[s]++;cnt++;    }    map<string,int>::iterator iter;    for(iter = tree.begin();iter != tree.end();iter++){cout<<iter->first;        printf(" %.4f\n",iter->second*100.0/cnt);    }    return 0;}

Trie tree实现：

#include<iostream>#include<cmath>#include<cstdio>#include<cstring>#define MAXN 130using namespace std;struct Node{char name[35];int num;bool tree;Node *next[MAXN];Node(){num = 0;tree = false;for(int i=0;i<MAXN;i++){next[i] = NULL;}}};int sum;void insert(Node *root,char *str){int len = strlen(str);Node *q;q = root;for(int i=0;i<len;i++){int x = str[i];if(q->next[x] == NULL){q->next[x] = new Node;}q = q->next[x];}q->num++;q->tree = true;strcpy(q->name,str);}void output(Node *root){Node *q;q = root;if(q->tree){printf("%s %.4f\n",q->name,q->num*100.00/sum);}for(int i=0;i<MAXN;i++){if(q->next[i] != NULL){//q = q->next[i];output(q->next[i]);}}}int main(){//freopen("in.txt","w",stdin);//freopen("out.txt","r",stdout);Node *root = new Node;sum = 0;char str[35];while(gets(str) != 0){insert(root,str);//cout<<"****"<<str<<endl;sum++;}output(root);return 0;}

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