Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 6109 | Accepted: 2476 |
Description
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1
N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.
Input
Output
Sample Input
7 4 3 1 -4 2 8 -1000 2 1 2 6 6
Sample Output
3 3 1 2
Source
這個題目我剛開始設想不用兩個優先隊列來搞,只保存1個值之類的。後來發現把自己繞進去了。以後當思路明確且預估可以過的時候,就不要做無意義的優化吧。
left_pq這個優先隊列,表示當前box中,最小的i-1個數字(i只有在調用get的時候才增加,因此get指令的時候只push,而add指令時候,如果有push則必有pop)。
right_pq這個優先隊列,保存着第i小往後的所有數字。需要注意的是,這個堆需要重載比較符,它是一個小根隊列。
這樣的話,當遇到add指令,則需要與left_pq的最大元素比較,如果比之小,則替換之。將替換出來的數字再push回right_pq中。如果比之大,則直接push到right_pq中。
當遇到get指令時,則直接取right_pq的最小元素,然後pop之。再push到left_pq中。
如此做即可AC
提交記錄:
1、Accepted!
代碼:
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define oo 0x3f3f3f3f
#define MAXV 500 * 2
using namespace std;
int getv[30010];
int data[30010];
struct myint {
int num;
};
bool operator < (const myint & a, const myint & b) {
return a.num > b.num;
}
priority_queue pq_right;
priority_queue pq_left;
int main() {
int n, m;
scanf("%d%d", &m, &n);
int i, j, k;
for (i = 0; i < m; i++) scanf("%d", data+i);
for (i = 0; i < n; i++) scanf("%d", getv+i);
j = 0;
int last = -oo;
for (i = 0; i < m; i++) {
int left_max;
if (!pq_left.empty() && (left_max = pq_left.top()) && data[i] < left_max) {
pq_left.pop();
pq_right.push((myint){left_max});
pq_left.push(data[i]);
}
else {
pq_right.push((myint){data[i]});
}
while (i+1 == getv[j]) {
myint cur = pq_right.top();
pq_right.pop();
printf("%d\n", cur.num);
pq_left.push(cur.num);
j++;
}
}
return 0;
}