HDU 1053 Entropy 哈夫曼树

题意：根据哈夫曼编码原则压缩一个字符串，问压缩后的大小和压缩比。

思路：就是写一棵哈夫曼树。

http://acm.hdu.edu.cn/showproblem.php?pid=1053

ps:第一次写哈夫曼树，写得好丑

/*********************************************
    Problem : HDU 1053
    Author  : NMfloat
    InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++)
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;

int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};

char s[100005];
int a[30];
int tot = 0;

struct leaf {
    int x,y;
    int zero;
    char c;
    int num;
}t[1000];

struct Node {
    int num;
    int addr;
    Node(int _num,int _addr) {
        num = _num ; addr = _addr ;
    }
    bool operator < (const Node B) const {
        return num > B.num;
    }
};

priority_queue<Node>p;

int ans ;

void dfs(int u,int depth) {
    if(t[u].x == 0) {
        ans += depth * t[u].num;
        //printf("%c %d %d\n",t[u].c,depth,t[u].num);
        return ;
    }
    dfs(t[u].x,depth+1);
    dfs(t[u].y,depth+1);
}

void Huffman() {
    while(!p.empty())p.pop();
    rep(i,0,25) {
        if(a[i]) {
            //printf("%d ",a[i]);
            p.push(Node(a[i],tot));
            t[tot].x = t[tot].y = 0;
            t[tot].c = i + 'A';
            t[tot].num = a[i];
            tot ++;
        }
    }

    if(a[26]) {
        p.push(Node(a[26],tot));
        t[tot].x = t[tot].y = 0;
        t[tot].c = '_';
        t[tot].num = a[26];
        tot ++;
    }
    //puts("");
    if(p.size() == 1) {
        ans = strlen(s);
        return ;
    }
    while(p.size() != 1) {
        Node tmp1 = p.top();p.pop();
        Node tmp2 = p.top();p.pop();
        //printf("%d %d\n",tmp1.num,tmp2.num);
        t[tmp1.addr].zero = 0; t[tmp2.addr].zero = 1;

        t[tot].x = tmp1.addr ; t[tot].y = tmp2.addr;
        p.push(Node(tmp1.num+tmp2.num,tot));
        tot ++;
    }
    ans = 0;
    dfs(tot-1,0);
}

void input() {

}

void solve() {
    int len = strlen(s);
    cls(a,0);
    rep(i,0,len-1) {
        if(s[i] == '_') a[26] ++;
        else a[s[i]-'A'] ++;
    }    
    tot = 1;
    Huffman();
    printf("%d %d %.1lf\n",len*8,ans,(double)(len*8)/(double)ans);
}

int main(void) {
    //freopen("a.in","r",stdin);
    while(~scanf("%s",s)) {
        if(strcmp(s,"END") == 0) break;
        input();
        solve();
    }
    return 0;
}