关于Huffman Codes的一道题
题目描述
题目来源及自己的思路
PTA - 中国大学MOOC-陈越、何钦铭-数据结构-2018秋
建树算最坏情况下的WPL值,然后根据输入建树,检查值是否在叶节点上,以及最终WPL值是否超过最坏情况。
相关代码
// 请把代码文本粘贴到下方(请勿用图片代替代码)
#include <stdio.h>#include <stdlib.h>
#include <string.h>
typedef struct TreeNode *HuffmanTree;
struct TreeNode {
HuffmanTree Left, Right;
int Weight;
char Data;
};
typedef struct HeapNode *MinHeap;
struct HeapNode {
HuffmanTree data[64];
int size;
};
MinHeap ini_a_heap(int num) {
MinHeap the;
the = malloc(sizeof(struct HeapNode));
the -> size = num;
char ch;
int we;
for (int i = 0; i < num; i++) {
scanf(" %c %d", &ch, &we);
the->data[i] = malloc(sizeof(struct TreeNode));
the->data[i]->Data = ch;
the->data[i]->Weight = we;
the->data[i]->Left = NULL;
the->data[i]->Right = NULL;
}
return the;
}
HuffmanTree ini_tree() {
HuffmanTree this;
this = malloc(sizeof(struct TreeNode));
this->Left = NULL;
this->Right = NULL;
return this;
}
int is_a_huff(MinHeap heap, int num, int best_wpl) {
char ch, code[num], j;
HuffmanTree tree, this;
tree = ini_tree();
int wpl = 0, res = 1;
for (int i = 0; i < num; i++) {
int dep = 0;
scanf(" %c %s", &ch, code);
if (strlen(code) > num - 1)
res = 0;
printf("%d %d\n",strlen(code), num - 1);
this = tree;
if (res)
for (j = 0; j < strlen(code) - 1; j++) {
dep++;
switch (code[j]) {
case '0':
if (this->Left == NULL) {
this->Left = ini_tree();
}
this = this->Left;
break;
case '1':
if (this->Right == NULL) {
this->Right = ini_tree();
}
this = this->Right;
break;
}
}
if (res)
switch (code[j]) {
case '0':
if (this->Left == NULL)
this->Left = heap->data[i];
else
res = 0;
break;
case '1':
if (this->Right == NULL)
this->Right = heap->data[i];
else
res = 0;
break;
}
wpl += heap->data[i]->Weight * (dep+1);
}
if (wpl <= best_wpl && res) return 1;
return 0;
}
int com(const void *a, const void *b) {
return *(int *)b - *(int *)a;
}
int WPL(MinHeap heap, int num) {
int i, weight[num], depth = 0, res = 0;
for (i = 0; i < num; i++)
weight[i] = heap->data[i]->Weight;
qsort(weight, num, sizeof(int), com);
for (i = 0; i < num - 2; i++)
res += (++depth) * weight[i];
res += (++depth) * weight[i++];
res += depth * weight[i];
return res;
}
int main(int argc, char **argv) {
int N, wpl;
scanf("%d", &N);
MinHeap heap;
heap = ini_a_heap(N);
wpl = WPL(heap, N);
int time;
scanf("%d", &time);
for (int i = 0; i < time; i++) {
if (is_a_huff(heap, N, wpl)) puts("Yes");
else puts("No");
}
return 0;
}
你期待的结果是什么?实际看到的错误信息又是什么?
提交后:
最大N&M,code长度等于63,这个到底是什么意思,N最大63不就意味着code长度不应该达到63吗?怎么会wrong answer?
回答:
说一下我遇到的情况。
实现语言:c。
问题所在:对输入数据的接收有问题。
对输入数据“c[i] f[i]”,我尝试都用%c来接收,即scanf(" %c", &temp),当f[i]=1000时,输入数据并不能正确的被接收。最终的输出结果自然也有问题。
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