日期:2014-05-20  浏览次数:20808 次

ACM上的一道题,求代码
地址在这:http://ac.jobdu.com/problem.php?id=1172

题目描述:
哈夫曼树,第一行输入一个数n,表示叶结点的个数。需要用这些叶结点生成哈夫曼树,根据哈夫曼树的概念,这些结点有权值,即weight,题目需要输出所有结点的值与权值的乘积之和。
输入:
输入有多组数据。
每组第一行输入一个数n,接着输入n个叶节点(叶节点权值不超过100,2<=n<=1000)。
输出:
输出权值。
样例输入:
5  
1 2 2 5 9
样例输出:
37


我用java做出来了。但是感觉效率不高。
有没有不建树的做法??

------解决方案--------------------
C/C++ code
#include<iostream>
#include<cstdio>
using namespace std;
#include<stdlib.h>
#include<memory.h>

typedef struct
{
    int weight;
    int flag;
    int parent,lchild,rchild;
}Node,*HuffmanTree;       //动态分配数组存储赫夫曼树 
 
int L,length;

void Select(HuffmanTree HT, int num, int &s1, int &s2)
{
    //s1是最小值,s2是次小值
    int i,minx = 1<<30;
    int mminx = 1<<30;
    for(i = 1; i <= num; ++i)
    {
        if(HT[i].weight < minx && !HT[i].flag)
        {
            mminx = minx; s2 = s1;
            minx = HT[i].weight; s1 = i;
        }
        else if(HT[i].weight < mminx && !HT[i].flag)
        {
            mminx = HT[i].weight; s2 = i;
        }
    }
    HT[s1].flag = HT[s2].flag = -1;
}
 
void AllLength(HuffmanTree HT, int m)
{
    int m1,m2;
    length++;
    if (!HT[m].lchild && !HT[m].rchild)
    {
        L += HT[m].weight * (length-1);
    }
    else
    {
        m1 = HT[m].lchild;
        AllLength(HT,m1);
        length--;
        m2 = HT[m].rchild;
        AllLength(HT,m2);
        length--;
    }
}
 
 
void HuffmanCoding(HuffmanTree &HT,int *w, int n,int m)
{
    int i, s1,s2;
    //w存放n个字符的权值(均>0),构造赫夫曼树HT
    HT = (HuffmanTree)malloc((m+1)*sizeof(Node));  //0号单元未用 
    HuffmanTree p = HT + 1;
    for(i = 1; i <= n; ++i,++p)
    {
        p->weight = w[i];
        p->parent = p->lchild = p->rchild = p->flag = 0;
    }
    for(i = n+1; i <= m; ++i,++p)
    {
        p->weight = p->parent = p->lchild = p->rchild = p->flag = 0;
    }
    
    for(i = n + 1; i <= m; ++i)
    {   //建赫夫曼树 
        //在HT[1..i-1]选择parent为0且weight最小的两个结点,其序号分别为s1和s2。
        Select(HT, i-1, s1, s2); 
        HT[s1].parent = i; HT[s2].parent = i;
        HT[i].lchild = s1; HT[i].rchild = s2;
        HT[i].weight = HT[s1].weight + HT[s2].weight;
    }
}
 
int main(void)
{
    int i,n,m;
    int w[1010];
    while(scanf("%d",&n)!=EOF)
    {
        m = 2 * n - 1;
        HuffmanTree T;
        for(i = 1; i <= n; ++i)
        {
            scanf("%d",&w[i]);
        }
        HuffmanCoding(T,w,n,m);
        length = L = 0;
        AllLength(T,m);
        free(T);
        printf("%d\n",L);
    }
    return 0;
}