求哈夫曼树的平均码长,用优先队列来写,先记录某个字符在字符串里出现的次数,然后放入
队列。依次取出第一小和第二小的数,将两个数相加,构成新的虚拟结点,放入队列中。
/*Accepted 196K 0MS C++ 918B 2012-08-01 17:25:00*/ #include<stdio.h> #include<string.h> #include<stdlib.h> #include<queue> using namespace std; int key[1 << 7], len; char t[1 << 10]; int huffman() { int i, a, b, c, sum = 0; priority_queue<int, vector<int>, greater<int> > q; for(i = 65; i < 100; i ++) { if(key[i] > 0) q.push(key[i]); } while(q.size() > 1) { a = q.top(); q.pop(); b = q.top(); q.pop(); c = a + b; sum += c; q.push(c); } return sum > 0 ? sum : len; //当只有一个字符的时候 } int main() { int i, sum; while(scanf("%s", t) != EOF) { if(strcmp(t, "END") == 0) break; len = strlen(t); memset(key, 0, sizeof key); for(i = 0; i < len; i ++) key[t[i]] ++; sum = huffman(); printf("%d %d %.1f\n", len * 8, sum, (double)len * 8 / sum); } return 0; }