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连接(应该可以这么说吧)优先队列——POJ 1442
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Black Box
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
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
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.
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
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.
Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.
Sample Input
7 4 3 1 -4 2 8 -1000 2 1 2 6 6
Sample Output
3 3 1 2
题意:给定M个数,每次可以插入序列一个数;再给N个数,表示在插入第几个数时输出一个数,第一次输出序列中最小的,第二次输出序列中第二小的……以此类推,直到输出N个数。
分析:因为输出时是按照先输出最小的,再输出第二小这样的方式输出的,相当于依次输出一个有序序列中的值。但因为这个序列不是固定不变的,而是不断的在更新,所以用数组是无法实现的。我们可以用优先队列来做。
定义两个优先队列,一个用来存储前k小的数,大数在前,小数在后;另一个优先队列第k+1小到最大的数,小数在前,大数在后。每次拿到一个数,先判断第一个优先队列中的数满不满k个,如果不满k个,则直接把这个数压入到第一个队列;如果满k个,判断这个数和第一个优先队列中的第一个数的大小:如果比第一个数大,就压入第二个优先队列;如果比第一个数小,就把第一个优先队列的队首元素弹出压入第二个队列,把这个新数压入第一个优先队列。
别人的题解,颇具启发性~
两个优先队列即是这样:
#include<cstdio> #include<cstdlib> #include<cmath> #include<map> #include<set> #include<queue> #include<stack> #include<vector> #include<algorithm> #include<cstring> #include<string> #include<iostream> const int MAXN=30000+10; using namespace std; int a[MAXN]; int b[MAXN]; int main() { freopen("in.txt","r",stdin); int n,m; while(scanf("%d%d", &n,&m)==2) { priority_queue<int>q1;//栈顶元素最大 priority_queue<int,vector<int>,greater<int> >q2;//栈顶元素最小 int i,j; for(i=1; i<=n; i++){ scanf("%d", &a[i]); } for(i=1; i<=m; i++){ scanf("%d", &b[i]); } j=1; int k=1;//要输出第k小的数 for(i=1; i<=n; i++){//把数放入栈 if(q1.size()<k) q1.push(a[i]);//q1如果没有达到k个数,把数放入q1 else{ //否则,用a[i]和q1队首元素比较,小的在q1,大的在q2 int t=q1.top(); if(a[i]>=t) q2.push(a[i]); else{ q2.push(t); q1.pop(); q1.push(a[i]); } } while(i==b[j])//在放入第b[j]个数后就要输出第k小的数,即q1队首元素 { printf("%d\n",q1.top()); k++; j++; if(!q2.empty()){ //q1长度+1,从q2取最小元素放入q1,继续判断是不是第b[j]次放入数字 q1.push(q2.top()); q2.pop(); } } } } return 0; }
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