HDU 1053 && PKU 1521 Entropy (PKU 3253, HDU 2527同解)

Huffman编码， 题意：给你一个序列，output the length in bits of the 8-bit ASCII encoding, the length in bits of an optimal prefix-free variable-length encoding, and the compression ratio accurate to one decimal point.

 

其实就是要你构造最优前缀树， 也就是Huffman树， 求它的带权路径和。 刚开始每种字符代表一个根节点， 它出现的次数就是其权值， 接下来就是求构造的Huffman树的带权路径和了。

 

刚开始比较傻， 先建立了Huffman树的二叉链表，然后通过先序遍历求它的带权路径和。

 

code：

24963902010-05-28 12:34:39Accepted10530MS256K1670 BG++xwc-hdu

#include <cstdio>#include <cstring>using namespace std;char str[1010];bool hash[1010];int num[220];typedef struct BitNode{ int data; BitNode *lchild, *rchild; BitNode() { lchild = NULL; rchild = NULL; }}*BitTree;BitTree T[1010];int sum;int Min(BitTree T[], int k)//求最小权值的根，返回根的下标 { int min, i; T[k] = new BitNode; T[k]->data = 0x3fffffff; min = k; for (i = 0; i < k; i++) { if (!hash[i] && T[i]->data < T[min]->data) min = i; } hash[min] = true; return min;}void Allsum(BitTree &T, int num)//先序求带权路径和 { if (T) { num++;//num表示该路径上经过的节点个数 if (!T->lchild && !T->rchild) { sum += (num - 1) * T->data; return ; } Allsum(T->lchild, num); Allsum(T->rchild, num); }}int main(){ int k, i, ans1; while (scanf("%s", str) != EOF) { if (strcmp(str, "END") == 0) break; memset(num, 0, sizeof(num)); memset(hash, false, sizeof(hash)); for (i = 0; str[i]; i++) { num[str[i]]++; } k = 0; for (i = 0; i < 150; i++) { if (num[i] != 0) { T[k] = new BitNode; T[k++]->data = num[i]; } } if (k == 1)//即表示只出现一个字符，最短编码长度即字符串长度 { ans1 = strlen(str) * 8; sum = strlen(str); printf("%d %d %.1lf/n", ans1, sum, ans1 * 1.0/sum); continue; } int len = k; while (--len) { int tmp1 = Min(T, k); int tmp2 = Min(T, k); T[k] = new BitNode; T[k]->lchild = T[tmp1]; T[k]->rchild = T[tmp2]; T[k]->data = T[tmp1]->data + T[tmp2]->data; k++; } BitTree root = new BitNode; root = T[k-1]; sum = 0; Allsum(root, 0); ans1 = strlen(str) * 8; printf("%d %d %.1lf/n", ans1, sum, ans1 * 1.0/sum); } return 0;} 

后来发现不用建树的， 因为是求树的带权路径长度， 根据权值可以直接求得。

#include <cstdio>#include <cstring>#include <queue>#include <vector>using namespace std;int main(){char str[1010];int i, test, k, nn, m, current, sum, num[150];priority_queue<int, vector<int>, greater<int> >Q;while (scanf("%s", str) != EOF){if (strcmp(str, "END") == 0)break;while (!Q.empty())Q.pop();memset(num, 0, sizeof(num));for (i = 0; str[i]; i++){num[str[i]]++;}k = 0;for (i = 'A'; i <= 'Z'; i++){if (num[i] != 0){Q.push(num[i]);k++;}}if (num['_'] != 0){Q.push(num['_']); k++;}nn = k;sum = 0;while (--k){current = Q.top();Q.pop();current += Q.top();Q.pop();sum += current;Q.push(current);}if (nn == 1)sum = strlen(str);int ans1 = strlen(str) * 8; printf("%d %d %.1lf/n", ans1, sum, ans1 * 1.0/sum);}return 0;}