poj 2418 Hardwood Species

http://poj.org/problem?id=2418

题意:给定一些树,按字典序输出数名和树出现的频率;

思路:这个题用二叉搜索树可以做,同时在网上看到了一个简单的方法,用map来做,这里map真是太好用了;

map做法:

View Code
#include <iostream>
#include <map>
#include <cstdio>
#include <string>
using namespace std;
int main()
{
string a;
map<string,int>tree;
double n = 0;
char ss[35] = {NULL};
while(gets(ss))
{
a = ss;
++tree[a];
++n;
}
map<string,int>::iterator temp;
for(temp = tree.begin(); temp != tree.end(); ++temp)
{
cout<<temp->first<<" ";
printf("%.4f\n",(temp->second)*100.0/n);
}
return 0;
}

二叉搜索树:

刚开始用stirng存名字,G++就是tle而c++就是CE ,无语了;

代码:

View Code
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
struct node
{
char s[40];
int n;
node *left,*right;
};
int n = 0;
node *get()
{
node *tt = NULL;
tt = new node;
tt->n = 1;
tt->left = NULL;
tt->right = NULL;
return tt;
}
node *head = NULL;
node *temp = NULL;
char str[40] = {NULL};
void insert(char *str,node *t)
{
if(strcmp(str,t->s) == 0)
++t->n;
else
if(strcmp(str,t->s) < 0)
{
if(t->left == NULL)
{
t->left = get();
t->left->n = 1;
strcpy(t->left->s,str);
return ;
}
else
insert(str,t->left);
}
else
{
if(t->right == NULL)
{
t->right = get();
t->right->n = 1;
strcpy(t->right->s,str);
return ;
}
else
insert(str,t->right);
}
}
void output(node *tt)
{
if(tt->left == NULL && tt->right == NULL)
{
printf("%s %.4f\n",tt->s,tt->n*100.00/n);
return ;
}
if(tt->left)
output(tt->left);
printf("%s %.4f\n",tt->s,tt->n*100.00/n);
if(tt->right)
output(tt->right);
}
int main()
{
//freopen("input.txt","r",stdin);
gets(str);
head = get();
strcpy(head->s,str);
head->n = 1;
//puts(head->s);
n = 1;
while(gets(str) != NULL)
{
++n;
insert(str,head);
}
//printf("%d\n",n);
output(head);
return 0;
}




posted @ 2012-03-23 19:26  LT-blogs  阅读(188)  评论(0编辑  收藏  举报