題干：

The only printer in the computer science students' union is experiencing an extremely heavy workload. Sometimes there are a hundred jobs in the printer queue and you may have to wait for hours to get a single page of output.?

Because some jobs are more important than others, the Hacker General has invented and implemented a simple priority system for the print job queue. Now, each job is assigned a priority between 1 and 9 (with 9 being the highest priority,?
and 1 being the lowest), and the printer operates as follows.

• The first job J in queue is taken from the queue.
• If there is some job in the queue with a higher priority than job J, thenmove J to the end of the queue without printing it.
• Otherwise, print job J (and do not put it back in the queue).

In this way, all those importantmuffin recipes that the Hacker General is printing get printed very quickly. Of course, those annoying term papers that others are printing may have to wait for quite some time to get printed, but that's life.?

Your problem with the new policy is that it has become quite tricky to determine when your print job will actually be completed. You decide to write a program to figure this out. The program will be given the current queue (as a list of priorities) as well as the position of your job in the queue, and must then calculate how long it will take until your job is printed, assuming that no additional jobs will be added to the queue. To simplifymatters, we assume that printing a job always takes exactly one minute, and that adding and removing jobs from the queue is instantaneous.

Input

One line with a positive integer: the number of test cases (at most 100). Then for each test case:

• One line with two integers n and m, where n is the number of jobs in the queue (1 ≤ n ≤ 100) and m is the position of your job (0 ≤ m ≤ n ?1). The first position in the queue is number 0, the second is number 1, and so on.
• One linewith n integers in the range 1 to 9, giving the priorities of the jobs in the queue. The first integer gives the priority of the first job, the second integer the priority of the second job, and so on.

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Output

For each test case, print one line with a single integer; the number of minutes until your job is completely printed, assuming that no additional print jobs will arrive.

Sample Input

3 1 0 5 4 2 1 2 3 4 6 0 1 1 9 1 1 1

Sample Output

1 2 5

題目大意：

一個打印序列，有優(yōu)先級別順序。從前往后掃描，若該作業(yè)優(yōu)先級別已是最高，則執(zhí)行此作業(yè)。若不是，放到隊列末尾。求指定位置的作業(yè)mine被執(zhí)行的時間。

解題報告：

? ? ? ?用一個隊列和優(yōu)先隊列來模擬整個過程就可以了，剛開始非要只用優(yōu)先隊列然后找個規(guī)律就可以了，結(jié)果發(fā)現(xiàn)有很多特殊情況需要考慮，最后放棄了這一種做法。

AC代碼：

#include <iostream> #include <cstdio> #include <queue> using namespace std; int n,vip,mine,res; int main() {int t;scanf("%d",&t);while (t--) {queue<int> q;priority_queue<int> pq;scanf("%d%d",&n,&mine);mine++;for (int i = 0; i < n; ++i) {scanf("%d",&vip);q.push(vip);pq.push(vip);}res = mine;while (1) {int cur = q.front();q.pop();if (res == 1) {//如果打印的是當(dāng)前任務(wù) if (cur != pq.top()) {//還有優(yōu)先級更高的。q.push(cur);res = pq.size();} else break;} else {res--;if (cur != pq.top()) {q.push(cur);}else pq.pop();}}printf("%d\n", n-q.size());}return 0; }

錯誤代碼：

#include<bits/stdc++.h> #define ll long long using namespace std; const int MAX = 1e2 + 5 ; int n,mine,myvip,myid; int bk[15]; struct Node {int id,vip;Node(){}Node(int id,int vip):id(id),vip(vip){}bool operator < (const Node b) const{if(vip != b.vip) return vip < b.vip;return id > b.id;} } node[MAX]; int main() {int t;cin>>t;while(t--) {priority_queue<Node> pq;cin>>n>>mine;mine++;for(int i = 1; i<=n; i++) {scanf("%d",&node[i].vip);node[i].id=i;pq.push(node[i]);bk[node[i].vip]++;if(i==mine) myvip = node[i].vip,myid = bk[myvip]-1;//myid記錄前面有幾個 } int ans = 0,tmp = 0;for(int i = 1; i<=n; i++) {if(pq.top().vip == myvip) break;pq.pop();ans++;}printf("ans = %d\n",ans);for(int i = n; i>=1; i--) {if(node[i].vip > myvip) break;if(node[i].vip == myvip) tmp++;}printf("tmp = %d\n",tmp); // while(!pq.empty()) { // if(pq.top().id == mine) break;pq.pop(); // tmp++; // }printf("%d\n",ans + myid + tmp + 1);}return 0 ; }

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