UVA 12676 Inverting Huffman

题目链接：https://vjudge.net/problem/UVA-12676

题目大意

　　一串文本中包含 N 个不同字母，经过哈夫曼编码后，得到这 N 个字母的相应编码长度，求文本的最短可能长度。

分析

　　哈夫曼树有这样一个性质，对于位于第 i 层的节点 A 和 第 i + 1 层的节点 B，A 出现的频率肯定大于等于 B，不然就可以把这两个节点互换，可以得到更优的哈夫曼树，因此，A 所能取到的最小频率，就是 i + 1 层所有节点出现频率的最大值，而根据题意，哈夫曼树的最底层节点频率可以全部设置为 1，如此可以一层层往上递推。

代码如下

 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef pair< int, int > PII;
 typedef pair< int, PII > PIPII;
 typedef pair< string, int > PSI;
 typedef pair< int, PSI > PIPSI;
 typedef set< int > SI;
 typedef vector< int > VI;
 typedef vector< VI > VVI;
 typedef vector< PII > VPII;
 typedef map< int, int > MII;
 typedef map< int, PII > MIPII;
 typedef map< string, int > MSI;
 typedef multimap< int, int > MMII;
 //typedef unordered_map< int, int > uMII;
 typedef pair< LL, LL > PLL;
 typedef vector< LL > VL;
 typedef vector< VL > VVL;
 typedef priority_queue< int > PQIMax;
 typedef priority_queue< int, VI, greater< int > > PQIMin;
 const double EPS = 1e-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e9 + ;
 const int maxN = 1e4 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 struct Node{
     LL weight = , len;

     bool operator< (const Node &x) const {
         if(len == x.len) return weight > x.weight;
         return len < x.len;
     }
 };

 int N; 

 int main(){
     //freopen("MyOutput.txt","w",stdout);
     //freopen("input.txt","r",stdin);
     INIT();
     while(cin >> N) {
         priority_queue< Node > maxH;
         Rep(i, N) {
             Node t;
             cin >> t.len;
             maxH.push(t);
         }

         // nowLen: 记录当前处理第几层
         // nowMax: 记录 i + 1 层的最大值
         // tmpMax: 记录当前层 i 层的最大值
         LL nowLen = maxH.top().len, nowMax = , tmpMax = -;
         while(maxH.size() > ) {
             Node a = maxH.top(); maxH.pop();
             Node b = maxH.top(); maxH.pop();

             if(a.len == nowLen) {
                 // 代码看到这里也许会有疑问，就是如果 a.weight == 0 或是 b.weight == 0，难道不要把节点扔回去重新取吗？
                 // 其实不需要，因为它们就是最小的
                 // 设max(i)表示倒数第 i 层的频率最大值，min(i)表示倒数第 i 层的频率最小值
                 // 容易看出 max(i) <= 2^i, min(i) >= 2^{i-1}
                 // 于是倒数第 i + 1 层权值为 0 的节点会被赋值为 max(i)
                 // 而那些权值不为 0 的节点，权值必然大于等于2倍的min(i)，为 2^i，大于等于 max(i)
                 if(a.weight == ) a.weight = nowMax;
                 if(b.weight == ) b.weight = nowMax;
                 tmpMax = max(tmpMax, b.weight);

                 a.weight += b.weight;
                 --a.len;
                 maxH.push(a);
             }
             else {
                 nowLen = a.len;
                 nowMax = tmpMax;
                 tmpMax = -;
                 maxH.push(a);
                 maxH.push(b);
             }
         }
         cout << maxH.top().weight << endl;
     }
     return ;
 }