ASC2 C Hyperhuffman 优先队列

题意：给你一个序列 ，让你求这个序列组成哈夫曼树的 WPL

解题思路：优先队列直接搞。因为数太大，用了非递归求解。

解题代码：

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 // File Name: c.cpp
 3 // Author: darkdream
 4 // Created Time: 2015年04月14日 星期二 18时35分42秒
 5
 6 #include<vector>
 7 #include<list>
 8 #include<map>
 9 #include<set>
10 #include<deque>
11 #include<stack>
12 #include<bitset>
13 #include<algorithm>
14 #include<functional>
15 #include<numeric>
16 #include<utility>
17 #include<sstream>
18 #include<iostream>
19 #include<iomanip>
20 #include<cstdio>
21 #include<cmath>
22 #include<cstdlib>
23 #include<cstring>
24 #include<ctime>
25 #include<queue>
26 #define LL long long
27 #define maxn 500005
28 using namespace std;
29 struct node{
30    LL v;
31    int num ;
32    node(){}
33    node(LL _v , int _num){
34         v = _v ;
35         num = _num;
36    }
37    bool operator < (const node &b )const {
38      return v > b.v;
39    }
40 };
41 priority_queue<node> qu ;
42
43 int n ;
44 int ch[maxn*4][2];
45 int que[maxn*4];
46 int deep[maxn*4];
47 LL a[maxn];
48 int tot;
49 int main(){
50     freopen("huffman.in","r",stdin);
51     freopen("huffman.out","w",stdout);
52     scanf("%d",&n);
53     for(int i = 1;i <= n;i ++){
54        scanf("%I64d",&a[i]);
55        qu.push(node(a[i],i));
56     }
57     tot = n;
58     while(qu.size()!= 1){
59         node ta,tb;
60         ta = qu.top();
61         qu.pop();
62         tb = qu.top();
63         qu.pop();
64         //printf("%I64d %d %I64d %d\n",ta.v,ta.num,tb.v,tb.num);
65         tot ++ ;
66         ch[tot][0] = ta.num;
67         ch[tot][1] = tb.num;
68         qu.push(node(ta.v+tb.v,tot));
69     }
70     LL ans = 0 ;
71     int head = 0 ;
72     int tail = 0 ;
73     deep[tot] = 0 ;
74     for(que[tail++] = tot ; head < tail ;head ++)
75     {
76         //printf("%d %d %d\n",que[head],ch[que[head]][0],ch[que[head]][1]);
77         //:printf("%d\n",deep[que[head]]);
78         if(ch[que[head]][0]){
79             que[tail] = ch[que[head]][0];
80             deep[ch[que[head]][0]] = deep[que[head]]  + 1;
81             tail ++ ;
82         }
83         if(ch[que[head]][1]){
84             que[tail] = ch[que[head]][1];
85             deep[ch[que[head]][1]] = deep[que[head]]  + 1;
86             tail ++ ;
87         }
88     }
89     for(int i = 1;i <= n;i ++)
90         ans += 1ll * deep[i] * a[i];
91     printf("%I64d\n",ans);
92 return 0;
93 }