poj 1862 Stripies/优先队列

原题链接：http://poj.org/problem?id=1862

简单题，贪心+优先队列主要练习一下stl大根堆

写了几种实现方式写成类的形式还是要慢一些。。。

手打的heap：

1：

 1 #include<cstdio>

 2 #include<cstdlib>

 3 #include<cmath>

 4 #include<iostream>

 5 class Solution{

 6 public:

 7     static const int Max_N = 110;

 8     int sz;

 9     double heap[Max_N];

10     inline void init(){

11         sz = 0;

12     }

13     inline void push(double x){

14         int i = sz++;

15         while (i > 0){

16             int p = (i - 1) >> 1;

17             if (heap[p] >= x) break;

18             heap[i] = heap[p];

19             i = p;

20         }

21         heap[i] = x;

22     }

23     inline double pop(){

24         double ret = heap[0], x = heap[--sz];

25         int i = 0;

26         while ((i << 1) + 1 < sz){

27             int a = (i << 1) + 1, b = (i << 1) + 2;

28             if (b < sz && heap[a] <= heap[b]) a = b;

29             if (heap[a] <= x) break;

30             heap[i] = heap[a];

31             i = a;

32         }

33         heap[i] = x;

34         return ret;

35     }

36 };

37 int main(){

38 #ifdef LOCAL

39     freopen("in.txt", "r", stdin);

40     freopen("out.txt", "w+", stdout);

41 #endif

42     int n;

43     double a, b;

44     Solution ans;

45     ans.init();

46     scanf("%d", &n);

47     while (n--){

48         scanf("%lf", &a);

49         ans.push(a);

50     }

51     while (ans.sz > 1){

52         a = ans.pop(), b = ans.pop();

53         ans.push(2 * sqrt(a * b));

54     }

55     printf("%.3lf", ans.pop());

56     return 0;

57 }

View Code

2：

 1 #include<cstdio>

 2 #include<cstdlib>

 3 #include<cmath>

 4 #include<iostream>

 5 struct Node{

 6     static const int Max_N = 110;

 7     int sz;

 8     double heap[Max_N];

 9     Node() :sz(0){}

10     inline void push(double x){

11         int i = sz++;

12         while (i > 0){

13             int p = (i - 1) >> 1;

14             if (heap[p] >= x) break;

15             heap[i] = heap[p];

16             i = p;

17         }

18         heap[i] = x;

19     }

20     inline double pop(){

21         double ret = heap[0], x = heap[--sz];

22         int i = 0;

23         while ((i << 1) + 1 < sz){

24             int a = (i << 1) + 1, b = (i << 1) + 2;

25             if (b < sz && heap[a] <= heap[b]) a = b;

26             if (heap[a] <= x) break;

27             heap[i] = heap[a];

28             i = a;

29         }

30         heap[i] = x;

31         return ret;

32     }

33 }ans;

34 int main(){

35 #ifdef LOCAL

36     freopen("in.txt", "r", stdin);

37     freopen("out.txt", "w+", stdout);

38 #endif

39     int n;

40     double a, b;

41     scanf("%d", &n);

42     while (n--){

43         scanf("%lf", &a);

44         ans.push(a);

45     }

46     while (ans.sz > 1){

47         a = ans.pop(), b = ans.pop();

48         ans.push(2 * sqrt(a * b));

49     }

50     printf("%.3lf", ans.pop());

51     return 0;

52 }

View Code

3：

 1 #include<cstdio>

 2 #include<cstdlib>

 3 #include<cmath>

 4 #include<iostream>

 5 const int Max_N = 110;

 6 double heap[Max_N];

 7 int sz = 0;

 8 void push(double x){

 9     int i = sz++;

10     while (i > 0){

11         int p = (i - 1) >> 1;

12         if (heap[p] >= x) break;

13         heap[i] = heap[p];

14         i = p;

15     }

16     heap[i] = x;

17 }

18 double pop(){

19     double ret = heap[0], x = heap[--sz];

20     int i = 0;

21     while ((i << 1) + 1 < sz){

22         int a = (i << 1) + 1, b = (i << 1) + 2;

23         if (b < sz && heap[a] <= heap[b]) a = b;

24         if (heap[a] <= x) break;

25         heap[i] = heap[a];

26         i = a;

27     }

28     heap[i] = x;

29     return ret;

30 }

31 int main(){

32 #ifdef LOCAL

33     freopen("in.txt", "r", stdin);

34     freopen("out.txt", "w+", stdout);

35 #endif

36     int n;

37     double a, b;

38 

39     scanf("%d", &n);

40     while (n--){

41         scanf("%lf", &a);

42         push(a);

43     }

44     while (sz > 1){

45         a = pop(), b = pop();

46         push(2 * sqrt(a * b));

47     }

48     printf("%.3lf", pop());

49     return 0;

50 }

View Code

4：

std::priority_queue<double> ans

 1 #include<cstdio>

 2 #include<cstdlib>

 3 #include<cmath>

 4 #include<queue>

 5 #include<iostream>

 6 int main(){

 7 #ifdef LOCAL

 8     freopen("in.txt", "r", stdin);

 9     freopen("out.txt", "w+", stdout);

10 #endif

11     int n;

12     double a, b;

13     scanf("%d", &n);

14     std::priority_queue<double> ans;

15     while (n--){

16         scanf("%lf", &a);

17         ans.push(a);

18     }

19     while (ans.size() > 1){

20         a = ans.top(), ans.pop();

21         b = ans.top(), ans.pop();

22         ans.push(2 * sqrt(a * b));

23     }

24     printf("%.3lf", ans.top());

25     return 0;

26 }

View Code