2014-05-06 13:34

题目链接

原题:

we have a random list of people. each person knows his own height and the number of tall people in front of him. write a code to make the equivalent queue. 
for example : 
input: <"Height","NumberOfTall","Name">, 
<6,2,"A">,<1,4,"B">,<11,0,"C">,<5,1,"D">,<10,0,"E">,<4,0,"F"> 
output: "F","E","D","C","B","A" --> end of queue

题目:有一队人正在排队,告诉你每个人的姓名、身高(目前默认都不相同),以及他/她所看到的前面比他高的人。请你计算出这个队伍里各个人的顺序。

解法:从最矮的人入手。对于最矮的人,他所看到的所有人都比他高,所以如果他看到有K个人比他高,那么他一定排在K + 1名。我的代码是当时琢磨了很久,采用树状数组写出来的。此处的树状数组提供快速修改区间、查询单个元素的能力,能做到对数时间。现在居然已经忘了当时的具体思路。总体思想是先按身高升序排序,然后逐个确定每个人在队列里的位置。关键的一点:每次都只能确定当前最矮的人排在哪儿。树状数组中维护的值c[i]的意义,是记录当前位置的前面有多少个比它高。其中还有个方法用到二分搜索,所以整体复杂度是比较奇怪的O(n * log(n) +  n * log^2(n)) = O(n * log^2(n))。

代码:

  1 // http://www.careercup.com/question?id=4699414551592960
  2 #include <algorithm>
  3 #include <iostream>
  4 #include <string>
  5 #include <vector>
  6 using namespace std;
  7 
  8 struct Person {
  9     int height;
 10     int taller;
 11     string name;
 12     Person(int _height = 0, int _taller = 0, string _name = ""): 
 13         height(_height), taller(_taller), name(_name) {};
 14     bool operator < (const Person &other) {
 15         return this->height < other.height;
 16     };
 17     
 18     friend ostream& operator << (ostream &cout, const Person &p) {
 19         cout << '<'<< p.name << p.height << '>';
 20         return cout;
 21     };
 22 };
 23 
 24 template <class T>
 25 class BinaryIndexedTree {
 26 public:
 27     BinaryIndexedTree(int _n = 0): n(_n), v(_n + 1) {};
 28     
 29     int size() {
 30         return n;
 31     };
 32     
 33     void addAll(int x, const T &val) {
 34         while (x >= 1 && x <= n) {
 35             v[x] += val;
 36             x -= lowBit(x);
 37         }
 38     };
 39     
 40     void addInterval(int x, int y, const T &val) {
 41         addAll(x - 1, -val);
 42         addAll(y, val);
 43     };
 44     
 45     void clear() {
 46         v.resize(1);
 47         n = 0;
 48     };
 49     
 50     void resize(int new_n) {
 51         v.resize(new_n + 1);
 52         n = new_n;
 53     }
 54     
 55     T sum(int x) {
 56         T res = 0;
 57         while (x >= 1 && x <= n) {
 58             res += v[x];
 59             x += lowBit(x);
 60         }
 61         
 62         return res;
 63     };
 64     
 65     int lowerBound(const T &val) {
 66         int ll, mm, rr;
 67         
 68         if (n == 0) {
 69             return 0;
 70         }
 71 
 72         T res;
 73         if (val <= (res = sum(1))) {
 74             return 1;
 75         }
 76         if (val > (res = sum(n))) {
 77             return n + 1;
 78         }
 79         
 80         ll = 1;
 81         rr = n;
 82         while (rr - ll > 1) {
 83             mm = (ll + rr) / 2;
 84             res = sum(mm);
 85             if (val > res) {
 86                 ll = mm;
 87             } else {
 88                 rr = mm;
 89             }
 90         }
 91         
 92         return rr;
 93     }
 94 private:
 95     vector<T> v;
 96     int n;
 97     
 98     int lowBit(int x) {
 99         return x & -x;
100     };
101 };
102 
103 int main()
104 {
105     vector<Person> people;
106     vector<int> queue;
107     BinaryIndexedTree<int> bit;
108     int i, j;
109     int n;
110     
111     while (cin >> n && n > 0) {
112         people.resize(n);
113         for (i = 0; i < n; ++i) {
114             cin >> people[i].height >> people[i].taller >> people[i].name;
115         }
116         sort(people.begin(), people.end());
117         bit.resize(n);
118         queue.resize(n);
119         for (i = 1; i <= n; ++i) {
120             bit.addInterval(i, n, 1);
121         }
122         for (i = 1; i <= n; ++i) {
123             j = bit.lowerBound(people[i - 1].taller + 1);
124             queue[j - 1] = i;
125             bit.addInterval(j, n, -1);
126         }
127         for (i = 0; i < n; ++i) {
128             cout << people[queue[i] - 1];
129         }
130         cout << endl;
131         
132         people.clear();
133         queue.clear();
134         bit.clear();
135     }
136     
137     return 0;
138 }

 

 posted on 2014-05-06 13:59  zhuli19901106  阅读(403)  评论(0编辑  收藏  举报