Srm480 div1 250 points

Okay, I admit that this question is very simple (it won't be ignored by STL children). If I read the wrong question, it will lead to success... Speechless...

 

Today's 250 is not very difficult. The main points are:

1. For string cutting, the handwritten cut function is relatively safe. Of course, the STL string stream can be directly cut by space.

2. Search for keywords. Well, here is a good technique. Use the red/black tree (SET) to maintain all the dangerous keywords.

3. The judgment condition is that the dangerous website is only used when the dangerous keywords exceed the specified value. Otherwise, no.

4. When a website is a dangerous website, all its keywords are dangerous. At this time, we need to add all the keywords to the red/black tree for management.

5. For the reason of 4, to scan n times (similar to the bellmanford algorithm), the order of addition is not the original order. Here, the ID is sorted at the end,

You can use a ing map to store subscript.

 

 

My code (core part ):

Set <string> st; <br/> Map <string, int> ID; <br/> bool use [60]; </P> <p> bool comp (const string & A, const string & B) <br/>{< br/> return ID [a] <ID [B]; <br/>}</P> <p> bool check (const string & Key, int num) <br/>{< br/> int Len = key. length (); <br/> string ret; <br/> vector <string> V; <br/> int CNT = 0, size = 0; </P> <p> for (INT I = 0; I <Len; I ++) <br/> {<br/> ret = ""; <br/> while (I <Len & Key [I]! = '') <Br/>{< br/> RET + = Key [I]; <br/> I ++; <br/>}< br/> v. push_back (RET); <br/>}< br/> size = v. size (); </P> <p> for (INT I = 0; I <size; I ++) <br/> If (St. find (V [I])! = ST. end () <br/>{< br/> CNT ++; <br/>}</P> <p> If (CNT> = num) <br/> {<br/> for (INT I = 0; I <size; I ++) <br/> St. insert (V [I]); <br/> return true; <br/>}< br/> else <br/> return false; <br/>}</P> <p> class internetsecurity <br/>{< br/> public: <br/> vector <string> determinewebsite (vector <string> Add, vector <string> key, vector <string> Dang, int num) <br/>{< br/> int n = add. size (); <br/> vector <string> V; </P> <p> memset (Use, 0, sizeof (use); <br/> ID. clear (); <br/> St. clear (); <br/> for (INT I = 0; I <n; I ++) <br/> ID [add [I] = I; <br/> int size = Dang. size (); <br/> for (INT I = 0; I <size; I ++) <br/> St. insert (Dang [I]); </P> <p> for (int K = 0; k <n; k ++) <br/> for (INT I = 0; I <n; I ++) <br/> If (! Use [I] & check (Key [I], num) <br/>{< br/> use [I] = true; <br/> v. push_back (add [I]); <br/>}</P> <p> sort (v. begin (), V. end (), comp); <br/> return V; <br/>}< br/> };

 

Conclusion: Reading questions carefully should not be difficult to understand. It is a bad phenomenon to hate reading questions recently.