huffman code

讲得不错：

11111

 

Poj,3253

//Huffman每次选最小的两个加和最小，最后结果最小。

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #include 
 6 
 7 #define ll long long
 8 #define N 20005
 9 using namespace std;
10 
11 int n;
12 class HTree
13 {
14     public :
15     HTree* left;
16     HTree* right;
17     int weight;
18     
19     HTree(){left = right = NULL; weight=0;}
20     HTree(HTree* l,HTree* r,int w){left = l;    right = r;    weight=w;}
21     ~HTree(){delete left; delete right;}
22 };
23 
24 class Compare_tree
25 {
26 public:
27     bool operator () (HTree* t1, HTree* t2)
28     {
29         return t1->weight> t2->weight;
30     }
31 };
32 
33 ll BuildTree(int *frequency,ll charge)
34 {
35     priority_queue,Compare_tree> QTree;
36     
37     for (int i=0;i)
38     {
39         if(frequency[i])
40             QTree.push(new HTree(NULL,NULL,frequency[i]));
41     }
42     
43     while (QTree.size()>1)
44     {
45         HTree* lc  = QTree.top();
46         QTree.pop();
47         HTree* rc = QTree.top();
48         QTree.pop();
49         charge+=lc->weight+rc->weight;
50         HTree* parent = new HTree(lc,rc,lc->weight+rc->weight);
51         QTree.push(parent);
52     }
53     return charge;
54 }
55 
56 int main()
57 {
58     int L;
59     int freq[N] = {0};
60     scanf("%d",&n);
61     for(int i=0;i)
62     {
63         scanf("%d",&L);
64         freq[i]=L;
65     }
66     ll result=BuildTree(freq,0);
67     printf("%lld\n",result);
68     return 0;
69 }

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 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 
 6 #define ll long long
 7 #define N 20005
 8 using namespace std;
 9 
10 int n;
11 struct Node
12 {
13     int w;
14     bool operator<(const Node& a)const
15     {
16         return w>a.w;
17     }
18 };
19 
20 int main()
21 {
22     priority_queue qu;
23     Node root;
24     scanf("%d",&n);
25     for(int i=0;i)
26     {
27         
28         scanf("%d",&root.w);
29         qu.push(root);
30     }
31     ll result=0;
32     while(qu.size()>1)
33     {
34         int l=qu.top().w;qu.pop();
35         int r=qu.top().w;qu.pop();
36         root.w=l+r;
37         result+=root.w;
38         qu.push(root);
39     }
40     printf("%lld\n",result);
41     return 0;
42 }

用Huffman的最简洁形式重写了

 zoj,2399

注意Huffman的数据可能很大，所以注意weight和最后的和都用longlong

Huffman编码的长度等于 w[i]*l[i]就是每次建一次树的跟节点的权值

 1 #include 
 2 #include 
 3 #include 
 4 
 5 using namespace std;
 6 
 7 typedef long long ll;
 8 
 9 struct node
10 {
11     ll w;
12     
13     bool operator<(const node& a)const
14     {
15         return w>a.w;
16     }
17 };
18 
19 
20 int main()
21 {
22     int size;
23     scanf("%d",&size);
24     while(size--)
25     {
26         int n;
27         priority_queue qp;
28         scanf("%d",&n);
29         for(int i=0;i)
30         {
31             int w;
32             cin>>w;
33             node now;
34             now.w=w;
35             qp.push(now);
36         }
37         ll ans=0;
38         while(qp.size()>1)
39         {
40             ll l=qp.top().w;qp.pop();
41             ll r=qp.top().w;qp.pop();
42             
43             node now1;
44             now1.w=l+r;
45             qp.push(now1);
46             
47             ans+=now1.w;
48         }
49         
50         printf("%lld\n",ans);
51         if(size)
52             printf("\n");
53     }
54     return 0;
55 }

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poj,1521

学校系统真是不敢恭维，我感觉在针对我。。。。

交了一个小时都wa，我之后无可奈何跑到poj上去交，一下就过了。。我真的见了鬼了

 1 #include 
 2 #include 
 3 #include <string>
 4 #include 
 5 #include 
 6 #include 
 7 
 8 using namespace std;
 9 const int maxn=256;
10 struct Node
11 {
12     int weight;
13     
14     bool operator<(const Node& a)const
15     {
16         return weight>a.weight;
17     }
18 };
19 
20 
21 int main()
22 {
23     char s[maxn];
24     
25     while(gets(s))
26     {
27         if(!strcmp(s,"END"))
28             break;
29         int freq[maxn]={0};
30         int cnt=strlen(s);
31         priority_queue pq;
32         for(int i=0;i)
33         {
34             freq[s[i]]++;
35         }
36         for(int i=0;i<200;i++)
37         {
38             Node now;
39             if(freq[i])
40             {
41                 now.weight=freq[i];
42                 pq.push(now);
43                 freq[i]=0;
44             }
45         }
46         
47         int ans=0;
48         while(pq.size()>1)
49         {
50             int l=pq.top().weight;pq.pop();
51             int r=pq.top().weight;pq.pop();
52             
53             Node now;
54             now.weight=l+r;
55             pq.push(now);
56             ans+=now.weight;
57         }
58         if(!ans)
59             ans=cnt;
60         while(!pq.empty())
61             pq.pop();
62         printf("%d %d %.1lf\n",cnt*8,ans,double((double)(8*cnt)/ans));
63     }
64     return 0;
65 }

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转载于:https://www.cnblogs.com/do-it-best/p/5433011.html