传统_POJ-1442

题目链接：http://poj.org/problem?id=1442

堆的练手题。搞一个大根堆和一个小根堆，其余YY一下就好

 1 #include<cstdio>

 2 #include<cstdlib>

 3 #include<cstring>

 4 #include<algorithm>

 5 using namespace std;

 6 #define lowbit(a) ((a)&(-a))

 7 #define max(a, b) ((a)>(b)?(a):(b))

 8 #define min(a, b) ((a)<(b)?(a):(b))

 9 #define MAXN 30010

10 #define PI 3.1415926

11 #define E 2.718281828

12 #define INF 0x777777f

13 typedef long long LL;

14 

15 struct heap{

16     #define hMAXN 30010

17     int hCnt, a[hMAXN];

18     inline void clear(){hCnt = 0;}

19     inline void swp(int p, int q){a[q] = a[p];}

20     void swim(int p){

21         int tmp[] = {a[p]};

22         for (int q = p >> 1; q > 0; p = q, q >>= 1){

23             if (a[q] > tmp[0]){

24                 swp(q, p);

25             } else break;

26         } a[p] = tmp[0];

27     }

28     void sink(int p){

29         int tmp[] = {a[p]};

30         for (int q = p << 1; q <= hCnt; p = q, q <<= 1){

31             if (a[q] > a[q+1] && q < hCnt) q++;

32             if (a[q] < tmp[0]){

33                 swp(q, p);

34             } else break;

35         } a[p] = tmp[0];

36     }

37     void push(int x){a[++hCnt] = x; swim(hCnt);}

38     int pop(int p){int ret = a[p]; swp(hCnt, p); a[hCnt] = 0; hCnt--; swim(p); sink(p); return ret;}

39 } h1, h2;

40 int n, m, a[MAXN], b[MAXN];

41 

42 int main(){

43     scanf("%d%d", &n, &m);

44     for (int i = 1; i <= n; i++)

45         scanf("%d", &a[i]);

46     for (int i = 1; i <= m; i++)

47         scanf("%d", &b[i]);

48     int t = 1;

49     for (int i = 1; i <= n; i++){

50         if (a[i] < -h1.a[1] && h1.hCnt){

51             int tmp = h1.pop(1);

52             h1.push(-a[i]);

53             h2.push(-tmp);

54         }

55         else {

56             h2.push(a[i]);

57         }

58         for ( ; i == b[t]; t++){

59             printf("%d\n", h2.a[1]);

60             h1.push(-h2.pop(1));

61         }

62     }

63     return 0;

64 }