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UVa 10226 - Hardwood Species

题目:有很多不同名称的树,统计每种树出现的概率。

分析:字符串,字典树(trie)。直接利用字典树计数,然后排序输出即可。

说明:POJ2418没有测试组数,TLE几次才发现╮(╯▽╰)╭。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>

using namespace std;

char  words[32];

/* Trie define */  
typedef struct node0
{
	char name[32];
	int  times;
}tree;
tree T[12001];

bool cmp( tree a, tree b ) 
{
	return strcmp(a.name, b.name)<0;
}

#define nodesize 30003     //节点个数   
#define dictsize 128        //字符集大小   

typedef struct node1
{  
    int    flag;            //值域   
    node1* next[dictsize];  
}tnode;  
tnode  dict[nodesize];

class Trie  
{  
    private:
        int    size;  
        int    count;
        tnode* root;  
    public:  
        Trie() {initial();}  
        void initial() {  
            memset( dict, 0, sizeof( dict ) );  
            count = 0; size=0; root=newnode();  
        }  
        tnode* newnode() {return &dict[size ++];}  
        void insert( char* word ) {  
            tnode* now = root;  
            for ( int i = 0 ; word[i] ; ++ i ) {
                if ( !now->next[word[i]] )  
                    now->next[word[i]] = newnode();  
                now = now->next[word[i]];  
            }
            if ( !now->flag ) {
				now->flag = ++ count;
				strcpy(T[now->flag-1].name, word);
				T[now->flag-1].times = 1;
			}else T[now->flag-1].times ++;
        }
        void output(int sum) {
			sort(T, T+count, cmp);
			for ( int i = 0 ; i < count ; ++ i )
				printf("%s %.4lf\n",T[i].name,100.0/sum*T[i].times);
		}
}trie;  
/* Trie  end */  

int main()
{
	int t,count;
	scanf("%d",&t);
	getchar();
	getchar();
	while (t --) {
		trie.initial();
		count = 0;
		while ( gets(words) && strlen(words) ) {
			trie.insert( words );
			count ++;
		}
		trie.output(count);
		if (t) printf("\n");
	}
	return 0;
}

UVa 10226 - Hardwood Species