Sicily 3703. Huffman Coding V1

Huffman树……自己查了下资料才知道具体是怎么弄出来的，基本步骤是：
1.在节点集合中找出权值最小的两个节点，小者居左大者居右，并从集合中删除；
2.将上面找到的两个节点合二为一（权值），添加到节点集合；
3.重复上述步骤直到只剩下一个节点，此节点即为根节点；
4.查找某一个节点的编码时，从根节点出发，往左是0，往右是1，直到找到这个节点（必为叶子），整个历程就是这个节点的编码。
其实Huffman树并不是唯一的，因为若两个节点的权值相等时，左右子树并不确定。这道题中没有说明这种情况如何处理，因为数据并不会出现这样的情况（已测试），大家爱怎么处理就怎么处理。

Run Time: 0sec

Run Memory: 312KB

Code length: 1627Bytes

Submit Time: 2012-01-05 01:13:13

// Problem#: 3703
// Submission#: 1174156
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include 
#include 
#include 
using namespace std;

struct Node {
    string s;
    int weight;
    Node( string code, int n ) { s = code; weight = n; }
};

vector v;
vector::iterator it;

void encode( string s, int n ) {
    for ( it = v.begin(); it != v.end(); it++ ) {
        if ( n >= it->weight )
            break;
    }
    v.insert( it, Node( s, n ) );
}

int main()
{
    int n;
    int i;
    int keep;
    char c;
    string s, left, right;
    int freq[ 26 ] = { };
    string code[ 26 ] = { };

    cin >> n;
    while ( n-- ) {
        cin >> c;
        freq[ c - 'A' ]++;
    }

    for ( i = 0; i < 26; i++ ) {
        if ( freq[ i ] != 0 ) {
            s = 'A' + i;
            encode( s, freq[ i ] );
        }
    }

    while ( v.size() != 1 ) {
        left = v.back().s;
        for ( i = 0; i < left.size(); i++ )
            code[ left[ i ] - 'A' ] = '0' + code[ left[ i ] - 'A' ];
        keep = v.back().weight;
        v.pop_back();

        right = v.back().s;
        for ( i = 0; i < right.size(); i++ )
            code[ right[ i ] - 'A' ] = '1' + code[ right[ i ] - 'A' ];
        keep += v.back().weight;
        v.pop_back();

        encode( left + right, keep );
    }

    while ( true ) {
        keep = 0;
        for ( i = 1; i < 26; i++ ) {
            if ( freq[ i ] > freq[ keep ] )
                keep = i;
        }

        if ( freq[ keep ] != 0 ) {
            cout << (char)('A' + keep) << " " << freq[ keep ] << " " << code[ keep ] << endl;
            freq[ keep ] = 0;
        }
        else
            break;
    }

    return 0;

}