太心累了，主要是一开始建堆出现了小问题，导致测试点一直没过。心烦意乱，采用printf调试（bushi），终于找出来了，是因为child的范围没考虑，忘记限制2*i<N
简单记录下要点，吃饭去了：
哈夫曼编码要点
1. 编码长度之和一定等于哈夫曼编码的长度之和，也就是哈夫曼树各个叶结点的路径之和，所以采用哈夫曼树主要就是为了算出路径长度，图省事，我也没建立树，直接算总和。
2. 要满足前缀码都不相同，cyll在mooc上采用的是构建树的方法去看是否有字符落在不是叶节点上，我偷懒直接采用strcmp暴力遍历
满足这两点就可以算得上是最优编码了。
原题
In 1953, David A. Huffman published his paper "A Method for the Construction of Minimum-Redundancy Codes", and hence printed his name in the history of computer science. As a professor who gives the final exam problem on Huffman codes, I am encountering a big problem: the Huffman codes are NOT unique. For example, given a string "aaaxuaxz", we can observe that the frequencies of the characters 'a', 'x', 'u' and 'z' are 4, 2, 1 and 1, respectively. We may either encode the symbols as {'a'=0, 'x'=10, 'u'=110, 'z'=111}, or in another way as {'a'=1, 'x'=01, 'u'=001, 'z'=000}, both compress the string into 14 bits. Another set of code can be given as {'a'=0, 'x'=11, 'u'=100, 'z'=101}, but {'a'=0, 'x'=01, 'u'=011, 'z'=001} is NOT correct since "aaaxuaxz" and "aazuaxax" can both be decoded from the code 00001011001001. The students are submitting all kinds of codes, and I need a computer program to help me determine which ones are correct and which ones are not.
Input Specification:
Each input file contains one test case. For each case, the first line gives an integer?N?(2≤N≤63), then followed by a line that contains all the?N?distinct characters and their frequencies in the following format:
c[1] f[1] c[2] f[2] ... c[N] f[N]where?c[i]?is a character chosen from {'0' - '9', 'a' - 'z', 'A' - 'Z', '_'}, and?f[i]?is the frequency of?c[i]?and is an integer no more than 1000. The next line gives a positive integer?M?(≤1000), then followed by?M?student submissions. Each student submission consists of?N?lines, each in the format:
c[i] code[i]where?c[i]?is the?i-th character and?code[i]?is an non-empty string of no more than 63 '0's and '1's.
Output Specification:
For each test case, print in each line either "Yes" if the student's submission is correct, or "No" if not.
Note: The optimal solution is not necessarily generated by Huffman algorithm. Any prefix code with code length being optimal is considered correct.
Sample Input:
7
A 1 B 1 C 1 D 3 E 3 F 6 G 6
4
A 00000
B 00001
C 0001
D 001
E 01
F 10
G 11
A 01010
B 01011
C 0100
D 011
E 10
F 11
G 00
A 000
B 001
C 010
D 011
E 100
F 101
G 110
A 00000
B 00001
C 0001
D 001
E 00
F 10
G 11结尾无空行
Sample Output:
Yes
Yes
No
No结尾无空行
解答
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
//如果用struct的话可以真正实现哈夫曼树
//但这道题目其实只要求我们求出哈夫曼编码的长度，所以可以采用整型变量作数组类型
int N;
int char_weight[255];
int cnt;//the size of number
typedef struct Node* PtrNode;
typedef struct Node{
int data;
PtrNode left;
PtrNode right;
}Node;
typedef PtrNode Tree;
void Precdown(int heap[],int p){
int tmp=heap[p];
int i,child;
for(i=p;2*i<=N;i=child){
child=2*i;
if(child<N && heap[child+1]<heap[child])child++;
if(heap[child]<tmp)heap[i]=heap[child];
else break;
}
heap[i]=tmp;
}
void Precup(int heap[],int p){
int tmp=heap[p];
int i;
for(i=p;i>1;i/=2){
if(heap[i/2]>tmp){
heap[i]=heap[i/2];
}else break;
}
heap[i]=tmp;
}
int DeleteHeap(int heap[]){
int min=heap[1];
heap[1]=heap[N];
N--;
Precdown(heap,1);
return min;
}
void InsertHeap(int heap[],int x){
N++;
heap[N]=x;
Precup(heap,N);
}
void CreateMinHeap(int heap[]){
int i;
for(i=N/2;i>0;i--){
Precdown(heap,i);
}
}
int BuildHuff(int heap[]){
int length=0;
int v1,v2;
int root;
while(N>1){
v1=DeleteHeap(heap);
v2=DeleteHeap(heap);
root=v1+v2;
length+=root;
InsertHeap(heap,root);
}
return length;
}
int Judge(int length){
char c[100][100];
int total=0;
int num[255]={0};
int i=0,j=0;
int slength;
char ch;
int l1,l2;
for(i=0;i<cnt;i++){
scanf(" %c",&ch);
scanf(" %s",c[i]);
total+=char_weight[ch]*strlen(c[i]);
}
if(total!=length)return 0;//权值之和不同
//接下看有无重复的前缀码
else{
for(i=0;i<cnt;i<i++){
for(j=0;j<i;j++){
slength=(l1=strlen(c[i]))<(l2=strlen(c[j]))?l1:l2;
if(!strncmp(c[i],c[j],slength))return 0;
}
}
}
return 1;
}
int main(){
int i,j;
int weight;
char ch;
int ret;
int M;//the number of tests
scanf("%d",&N);
int heap[100];//start from 1
heap[0]=-1;
cnt=N;
for(i=0;i<N;i++){
scanf(" %c %d",&ch,&weight);
heap[i+1]=weight;
char_weight[ch]=weight;
}
CreateMinHeap(heap);
int length=BuildHuff(heap);
scanf("%d",&M);
for(i=0;i<M;i++){
ret=Judge(length);
if(ret)printf("Yes\n");
else printf("No\n");
}
return 0;
}
比较混乱，有闲情逸致再来修改总结。挥挥~
?