• hdu 1053 Entropy 霍夫曼树


    http://acm.hdu.edu.cn/showproblem.php?pid=1053

    /*
    **利用优先队列构造霍夫曼树,首先将所有编码出现次数作为叶子节点的权值加入队列,
    **然后每次取出权值最小的两个节点,组合后加入队列。
    **最后遍历树求得总的编码长度。
    */
    #include <iostream>
    #include <queue>
    #include <string>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    using namespace std;
    struct node
    {
        int w;
        bool yezi;//是否是叶子节点
        node *lc,*rc;
        bool operator< (const node &t) const//重载操作符,设定优先级
        {
            return t.w < w;
        }
    }a,b,c,*plc,*prc;
    int arr[31];
    
    int cmp(void const *a,void const *b){
            return *(int *)b - *(int *)a;
    }
    
    priority_queue<node> Q;//优先队列
    
    void HTree()
    {
        a = Q.top(); Q.pop();
        while(!Q.empty())
        {
            b = Q.top(); Q.pop();
            plc = new node();
            prc = new node();
            plc->w = a.w;plc->yezi = a.yezi; plc->lc = a.lc; plc->rc = a.rc;
            prc->w = b.w;prc->yezi = b.yezi; prc->lc = b.lc; prc->rc = b.rc;
            c.w = a.w + b.w;
            c.yezi = false;
            c.lc = plc;
            c.rc = prc;
            Q.push(c);
            a = Q.top(); Q.pop();
        }
    }
    
    int LOT(node *T, int step)
    {
        if(T->yezi) {return step*T->w;}
        return LOT(T->lc,step+1)+LOT(T->rc,step+1);
    }
    int main(){
        string s;
        while(cin>>s,s!="END")
        {
            memset(arr, 0, sizeof(arr));
            while(!Q.empty()) Q.pop();
            int i,sum=0;
            for(i=0; i<s.size(); i++)
                arr[s[i]-'A']++;
            qsort(arr,31,sizeof(arr[0]),cmp);
            if(arr[1]==0)//只有一种编码
            {
                printf("%d %d 8.0
    ",8*arr[0],arr[0]);
                continue;
            }
            for(i=0; i<31&&arr[i]!=0; i++)
            {
                a.w = arr[i];
                a.yezi = true;
                a.lc = a.rc = NULL;
                Q.push(a);
            }
            HTree();
            sum = LOT(&a, 0);
            printf("%d %d %.1f
    ",8*s.size(),sum,(8.0*s.size())/sum);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/tobec/p/3341914.html
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