九度OJ-1172-哈夫曼树

九度OJ-1172-哈夫曼树

http://ac.jobdu.com/problem.php?pid=1172

哈夫曼树(Huffman tree)：给定n个权值作为n个叶子结点，构造一棵二叉树，若带权路径长度达到最小，称这样的二叉树为最优二叉树,即哈夫曼树

这题可以建树，也可以不建树。

建树：

#include<stdio.h>#include<string.h>#include<stdlib.h>#define maxvalue 0x7fffffffstruct huffman{    int weight;int parent,lchild,rchild;}list[5000];int main(){int n,m;int i,j;int ans;int x1,x2;int m1,m2;while(scanf("%d",&n)!=EOF){m=2*n-1;for(i=0;i<m;i++)list[i].parent=list[i].lchild=list[i].rchild=-1;for(i=0;i<n;i++)scanf("%d",&list[i].weight);ans=0;for(i=0;i<n-1;i++){x1=x2=0;m1=m2=maxvalue;for(j=0;j<n+i;j++){if(list[j].weight<m1&&list[j].parent==-1){x2=x1;m2=m1;x1=j;m1=list[j].weight;}else if(list[j].weight<m2&&list[j].parent==-1){x2=j;m2=list[j].weight;}}list[x1].parent=n+i;list[x2].parent=n+i;list[n+i].lchild=x1;list[n+i].rchild=x2;list[n+i].weight=list[x1].weight+list[x2].weight;ans+=list[n+i].weight;}printf("%d\n",ans);}return 0;}

不建树的简单一点：

#include<stdio.h>#include<string.h>#include<stdlib.h>int cmp(const void *a,const void *b){return *(int *)a-*(int *)b;}int main(){int n,i,ans;int a[5000];while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++)scanf("%d",&a[i]);ans=0;for(i=0;i<n-1;i++){qsort(a+i,n-i,sizeof(a[0]),cmp);ans+=(a[i]+a[i+1]);a[i+1]+=a[i];}printf("%d\n",ans);}return 0;}

依据字符出现概率，可以利用哈夫曼树构造哈夫曼编码，写了个简易的代码：

给出26个字母的频度，执行哈夫曼编码，输入1编码，输入2译码。

#include<stdio.h>#include<string.h>#include<stdlib.h>#define maxvalue 0x7fffffffstruct cam1{char c;char s[50];int len;}zimu[30];struct cam2{int weight;int parent,lchild,rchild;}list[60];void init(){int i;for(i=0;i<26;i++)zimu[i].c='a'+i;}void huffmantree(){int i,j;int x1,x2;int m1,m2;int pindu[26]={64,13,22,32,103,21,15,47,57,1,2,32,20,57,63,15,1,48,51,80,23,8,18,1,16,1};  //26个字母的频度for(i=0;i<60;i++)  list[i].parent=list[i].lchild=list[i].rchild=-1;for(i=0;i<26;i++)list[i].weight=pindu[i];for(i=0;i<25;i++){x1=x2=0;m1=m2=maxvalue;for(j=0;j<26+i;j++){if(list[j].weight<m1&&list[j].parent==-1){x2=x1;m2=m1;x1=j;m1=list[j].weight;}else if(list[j].weight<m2&&list[j].parent==-1){x2=j;m2=list[j].weight;}}list[26+i].lchild=x1;list[26+i].rchild=x2;list[x1].parent=list[x2].parent=26+i;list[26+i].weight=list[x1].weight+list[x2].weight;}}void huffmanbianma(){int i,j,start;char s[100];int p,c,k;for(i=0;i<26;i++){memset(s,0,sizeof(s));start=40;c=i;p=list[c].parent;while(p!=-1){if(list[p].lchild==c)s[start]='0';elses[start]='1';start--;c=p;p=list[c].parent;}for(k=0,j=start+1;j<=40;k++,j++)zimu[i].s[k]=s[j];zimu[i].s[k]='\0';zimu[i].len=k;}printf("26个字母的哈夫曼编码如下：\n");for(i=0;i<26;i++)printf("%c----->%s\n",'a'+i,zimu[i].s);}int main(){int i,j;int ans,len;char str[1000];int k1,k2,flag;printf("Welcome to Cambridgeacm\n");init();huffmantree();huffmanbianma();printf("输入1时执行编码\n");printf("输入2时执行译码\n");while(scanf("%d",&ans)!=EOF){switch(ans){    case 1:{                 printf("请输入一段字符串\n");                 scanf("%s",str); len=strlen(str); for(i=0;i<len;i++) printf("%s",zimu[str[i]-'a'].s); printf("\n"); printf("编码结束\n");}case 2:{     printf("请输入一段01代码\n"); scanf("%s",str);                 len=strlen(str); for(i=0;i<len;i++) { for(j=0;j<26;j++) { flag=1; for(k1=i,k2=0;k1<=i+zimu[j].len-1&&str[k1]!='\0';k1++,k2++) if(str[k1]!=zimu[j].s[k2]) { flag=0; break; } if(flag) { printf("%c",zimu[j].c); i+=(zimu[j].len-1); break; } } } printf("\n"); printf("译码结束\n");}}}return 0;}

来个截图吧