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

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
  9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 12  
 13 #define pr(x) cout << #x << " = " << x << "  "
 14 #define prln(x) cout << #x << " = " << x << endl
 15  
 16 #define LOWBIT(x) ((x)&(-x))
 17  
 18 #define ALL(x) x.begin(),x.end()
 19 #define INS(x) inserter(x,x.begin())
 20  
 21 #define ms0(a) memset(a,0,sizeof(a))
 22 #define msI(a) memset(a,inf,sizeof(a))
 23 #define msM(a) memset(a,-1,sizeof(a))
 24 
 25 #define MP make_pair
 26 #define PB push_back
 27 #define ft first
 28 #define sd second
 29  
 30 template<typename T1, typename T2>
 31 istream &operator>>(istream &in, pair<T1, T2> &p) {
 32     in >> p.first >> p.second;
 33     return in;
 34 }
 35  
 36 template<typename T>
 37 istream &operator>>(istream &in, vector<T> &v) {
 38     for (auto &x: v)
 39         in >> x;
 40     return in;
 41 }
 42  
 43 template<typename T1, typename T2>
 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 45     out << "[" << p.first << ", " << p.second << "]" << "
";
 46     return out;
 47 }
 48 
 49 inline int gc(){
 50     static const int BUF = 1e7;
 51     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 52     
 53     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 54     return *bg++;
 55 } 
 56 
 57 inline int ri(){
 58     int x = 0, f = 1, c = gc();
 59     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
 60     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 61     return x*f;
 62 }
 63  
 64 typedef long long LL;
 65 typedef unsigned long long uLL;
 66 typedef pair< double, double > PDD;
 67 typedef pair< int, int > PII;
 68 typedef pair< int, PII > PIPII;
 69 typedef pair< string, int > PSI;
 70 typedef pair< int, PSI > PIPSI;
 71 typedef set< int > SI;
 72 typedef vector< int > VI;
 73 typedef vector< VI > VVI;
 74 typedef vector< PII > VPII;
 75 typedef map< int, int > MII;
 76 typedef map< int, PII > MIPII;
 77 typedef map< string, int > MSI;
 78 typedef multimap< int, int > MMII;
 79 //typedef unordered_map< int, int > uMII;
 80 typedef pair< LL, LL > PLL;
 81 typedef vector< LL > VL;
 82 typedef vector< VL > VVL;
 83 typedef priority_queue< int > PQIMax;
 84 typedef priority_queue< int, VI, greater< int > > PQIMin;
 85 const double EPS = 1e-10;
 86 const LL inf = 0x7fffffff;
 87 const LL infLL = 0x7fffffffffffffffLL;
 88 const LL mod = 1e9 + 7;
 89 const int maxN = 1e4 + 7;
 90 const LL ONE = 1;
 91 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 92 const LL oddBits = 0x5555555555555555;
 93 
 94 struct Node{
 95     LL weight = 0, len;
 96     
 97     bool operator< (const Node &x) const {
 98         if(len == x.len) return weight > x.weight;
 99         return len < x.len;
100     }
101 };
102 
103 int N; 
104 
105 int main(){
106     //freopen("MyOutput.txt","w",stdout);
107     //freopen("input.txt","r",stdin);
108     INIT();
109     while(cin >> N) {
110         priority_queue< Node > maxH;
111         Rep(i, N) {
112             Node t;
113             cin >> t.len;
114             maxH.push(t);
115         }
116         
117         // nowLen: 记录当前处理第几层
118         // nowMax: 记录 i + 1 层的最大值
119         // tmpMax: 记录当前层 i 层的最大值 
120         LL nowLen = maxH.top().len, nowMax = 1, tmpMax = -1;
121         while(maxH.size() > 1) {
122             Node a = maxH.top(); maxH.pop();
123             Node b = maxH.top(); maxH.pop();
124             
125             if(a.len == nowLen) {
126                 // 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？
127                 // 其实不需要，因为它们就是最小的
128                 // 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
129                 // 容易看出 max(i) <= 2^i, min(i) >= 2^{i-1}
130                 // 于是倒数第 i + 1 层权值为 0 的节点会被赋值为 max(i)
131                 // 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)
132                 if(a.weight == 0) a.weight = nowMax;
133                 if(b.weight == 0) b.weight = nowMax;
134                 tmpMax = max(tmpMax, b.weight);
135                 
136                 a.weight += b.weight;
137                 --a.len;
138                 maxH.push(a);
139             }
140             else {
141                 nowLen = a.len;
142                 nowMax = tmpMax;
143                 tmpMax = -1;
144                 maxH.push(a);
145                 maxH.push(b);
146             }
147         }
148         cout << maxH.top().weight << endl;
149     }
150     return 0;
151 }

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