HDU 1029 Ignatius and the Princess IV - Problem A

// AKC++.cpp : 定义控制台应用程序的入口点。////#include "stdafx.h"#include <iostream>#include <cstdio>#include <string>#include <cmath>#define MAX 1000005using namespace std;/**题意:找出N(奇数)个数中, 个数大于(N-1）/2的数分析1(TLE): 显然这些数必有一个在(N+1)/2这个位置,   用第快拍的思想找出第K大,虽然复杂度接近O(n) 但是可惜超时了分析2(AC):  哈希的思想, 输入一个就记录数字的个数+1, 如果有数字个数超过一半就不用再记了,  这样就快的多了.  在没给数据范围的情况下, 数组记录可能会溢出, 字典树是个不错的选择~算法:   trie树利用重复前缀的思想,使得存储空间极大节约, 存取过程为O(len) , len为记录的长度; */int N;int a[MAX];bool cmp(const int &lowVal, const int &highVal){    return lowVal <= highVal;}int partition(int low, int high){    int key = a[low];    while(low < high){        while( low < high && cmp(key, a[high])){high--;}a[low] = a[high];        while( low < high && !cmp(key, a[low])){low ++;}a[high] = a[low];    }a[low] = key;///在结果位置放入key值    return low;}void qSort(int low, int high){    if(low < high){        int mid = partition(low, high);        qSort(low, mid);        qSort(mid + 1, high);    }}void find_kthMax(int low, int high, int k){    if(low < high){        int mid = partition(low, high);        int kk = high - mid + 1;        if(kk == k)                return ;        else if(kk < k)                 find_kthMax(low, mid - 1, k - kk);        else             find_kthMax(mid + 1, high, k);    }}///------------------trie-------///跟hash思想一样,  但是利用重复前缀的特点, 极大节约了空间const int trieSize = 10;//数字字典10个分叉char num[20];bool isFind;char ans[20];struct Node{    int times;    Node *ChildNode[trieSize];    Node( ){        times = 0;        for( int i = 0; i < trieSize; i++){            ChildNode[i] =  NULL;        }    }};void insertNum(char* num, int len, Node *father, int i){    int ii = num[i] - '0';    if(father->ChildNode[ii]==NULL){        father->ChildNode[ii] = new Node( );    }    if(len == 1){        father->ChildNode[ii]->times++;        if(father->ChildNode[ii]->times >= N/2 + 1){            isFind = true;            strcpy(ans, num);        }    }    else insertNum(num, len -1, father->ChildNode[ii], i + 1);}int main(){        while(cin>>N){        Node *root = new Node();///暂时不考虑空间的释放        isFind = false;        for( int i = 0; i < N; i++){            cin>>num;            if(!isFind)                insertNum(num, strlen(num), root, 0);        }        cout<<ans<<endl;    }    return 0;}