uvalive2088

题目的意思就是哈夫曼编码问题.

刚开始用建树时,一直出问题 .后来听别人说了一种不用建树的办法.

就是把每个节点的频数和所用编码长度保存下来.

每次两个节点相结合后,那么新节点相应的频数就是两个频数相加,而总的编码长度就为之前两个频数之和加上两个长度之和.

计算到最后一个时,它对应的频数便是我们要求的.

然后输出就好.

三个输出是如果每个字母都用八字节的编码长度(字符串长度乘8) ,我们刚才求的编码长度. 和两个的比值.

AC代码:

#include <cstdio>#include <cstring>#include <string>#include <queue>#include <iostream>using namespace std;struct node {int num;int len;bool friend operator < (const node& a, const node& b) {return a.num > b.num;}}a[27];string s;priority_queue<node> q;int main() {while (cin >> s && s != "END") {while (!q.empty())q.pop();memset(a, 0, sizeof(a));int l = s.size();for (int i = 0; i < l; i++)if (s[i] == '_')a[26].num++;elsea[s[i] - 'A'].num++;for (int i = 0; i < 27; i++)if (a[i].num)q.push(a[i]);node tmp, t1, t2;if (q.size() == 1) {printf("%d %d 8.0\n", l * 8, l);continue;}while (q.size() != 1) {t1 = q.top();q.pop();t2 = q.top();q.pop();tmp.num = t1.num + t2.num;tmp.len = tmp.num + t1.len + t2.len;q.push(tmp);}printf("%d %d %.1lf\n", l * 8, tmp.len, double(l * 8) / tmp.len);}return 0;}