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 }