map的运用

Advertising

• Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others)
• Total Submission(s): 40     Accepted Submission(s): 6

Description

Maybe you have seen sports live, game live, etc. But you must have never seen GDCPC live before! Yes, the GDCPC committee decides to live stream the contest on their websites.

You know, when there is a video, there is an ad. The website developers want to insert ads into the live stream. Since it's the first time to serve ads, the website developers just want to use the most standard way, VAST (Video Ad Serving Template).

Basically, VAST is a XML and the video player plays ads inside it. An example looks like:

<VAST version="2.0">

   <Ad id="1">

       <InLine>

           <Creative AdID="601364" />

       </InLine>

   </Ad>

   <Ad id="2">

       <Wrapper>

           <VASTAdTagURI>http://demo.ad.com/vast.xml</VASTAdTagURI>

       </Wrapper>

   </Ad>

</VAST>

Inside a VAST, there are two kinds of ads: inline ad and wrapper ad.

For an inline ad, it contains one ID for interpret.

For a wrapped ad, it contains an URL, which wrap another VAST.

Video player interprets the VAST as following:

1. Check the ads inside the VAST sequentially

2. When encounter an inline ad, video player plays the media file inside it immediately, then move on to next ad.

3. When encounter a wrapper ad, video player need to fetch the wrapped VAST by its URL. Then interprets the wrapped VAST recursively. After the interpretation is done, move on to next ad.

Now the problem is to interpret the given VAST and output all inline ads IDs in order of appearance during interpretation.

Input

The first line is an integer T ( T ≤ 50 ), indicates the number of test cases.

The first line of each test case contains one integer n indicates the number of VAST.

For each VAST:

1. First line contains an URL and an integer k, which indicates the number of ads in this VAST

2. Then k lines follow, each line could be either "id inline" or "id wrapper URL"

It's guaranteed that in each test case

1. All VASTs have different URLs;

2. Total number of ads (both inline and wrapper) among these VAST is less than 1000; the interpretation always succeeds.

Output

For each test case, output "Case #t: s", t represents the index of test cases (start from 1), s represents the list of IDs when interpreting the first VAST, use space to separate the IDs.

If there is no ID when interpreting, s should be empty.

Sample Input

2
2
http://demo.ad.com/vast1 2
id1 wrapper http://demo.ad.com/vast2
id2 inline
http://demo.ad.com/vast2 1
id3 inline
4
http://demo.ad.com/vast1 2
id1 wrapper http://demo.ad.com/vast2
id2 wrapper http://demo.ad.com/vast3
http://demo.ad.com/vast2 1
id3 wrapper http://demo.ad.com/vast4
http://demo.ad.com/vast3 2
id4 inline
id5 inline
http://demo.ad.com/vast4 1
id6 inline

Sample Output

Case #1: id3 id2
Case #2: id6 id4 id5

Source

GDCPC 2017

 

题目大意：一个网页里面的广告，遇到wrapper表示跳转，inline表示已经访问到该广告。要求输出看到广告的先后顺序。

这题主要是运用对map映射的应用。

#include<bits/stdc++.h>
using namespace std;

map<string,vector<string> > mpa;
map<string,bool> mpb;
string root[1006];
int n;
bool flag;

void dfs(string x)
{
    if(mpb[x]){
        if(flag) cout<< " ";
        flag=1;
        cout << x ;
    }
    int len=mpa[x].size();
    for(int i=0;i<len;i++){
        dfs(mpa[x][i]);
    }
}

int main()
{
    ios::sync_with_stdio(0); std::cin.tie(0);
    int T;
    cin >> T;
    for(int iii=1;iii<=T;iii++){
        mpa.clear();
        mpb.clear();

        cin >> n;
        string y,z,a;
        int m;
        for(int i=1;i<=n;i++){
            cin >> root[i] >> m;
            for(int j=1;j<=m;j++){
                cin >> y >> z;
                if(z[0]=='i'){
                    mpb[y]=true;
                    mpa[root[i]].push_back(y);
                } else{
                    cin >> a;
                    mpa[root[i]].push_back(a);
                }
            }
        }
        cout << "Case #" << iii << ": " ;
        flag=0;
        dfs(root[1]);
        cout << endl;
    }
    return 0;
}