ccf刷题记04

不知不觉，近两年的ccf前两题都刷完了呢

大概总结一下，第一题就是个简单水题，热身签到

第二题是个简单模拟，多数是对数据或自创数据结构排序的考察，细心分析内部关系就好了

第三题就是个大模拟了，一般条件考虑还比较多，细心才能成大事

这次就来三道大模拟吧

三道题都是个字符串处理题，string，用getline读取，那么先关一下cin同步吧

ios::sync_with_stdio(false);

201703-3 Markdown

思路：字符串处理，根据每行首元素判断标题、段落和无序列表，然后分别按要求处理

对于下划线和链接，需要单独写个函数特判，下划线是个全局变量，因为其嵌套性很强，而链接中很可能还要套链接，于是写成递归吧

这题细节点很多，6次尝试后终于拿了满分

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <stack>using namespace std;bool down;string dealstring(string s){    int len = s.size();    string re = "";    for(int pos=0;pos<len;pos++)    {        if(s[pos] == '_')        {            if(down)            {                down = false;                re += "</em>";            }            else            {                down = true;                re += "<em>";            }            continue;        }        if(s[pos] == '[')        {            pos++;            int st = pos;            int slen = 1;            while(s[pos]!= ']')            {                slen++;                pos++;            }            string text = dealstring(s.substr(st,slen-1));            pos++;            pos++;            st = pos;            slen = 1;            while(s[pos]!= ')')            {                slen++;                pos++;            }            string link = dealstring(s.substr(st,slen-1));            re += "<a href=\"" + link + "\">" + text + "</a>";            continue;        }        re += s[pos];    }    return re;}int main(){    string line;    bool star = false;    bool paragraph = false;    //freopen("in3.txt","r",stdin);    down = false;    while(getline(cin,line))    {        int len = line.size();        if(len==0)        {            if(star)            {                star = false;                cout << "</ul>" << endl;            }            if(paragraph)            {                paragraph = false;                cout << "</p>" << endl;            }            continue;        }        if(line[0]=='#')        {            if(star)            {                star = false;                cout << "</ul>" << endl;            }            if(paragraph)            {                paragraph = false;                cout << "</p>" << endl;            }            int num = 1;            int pos = 1;            while(pos <= len)            {                if(line[pos]=='#')                {                    num++;                }                else if(line[pos]!=' ')                {                    break;                }                pos++;            }            cout << "<h" << num << ">";            cout << dealstring(line.substr(pos,len-pos));            cout << "</h" << num << ">" << endl;        }        else if(line[0]=='*')        {            if(paragraph)            {                paragraph = false;                cout << "</p>" << endl;            }            if(!star)            {                star = true;                cout << "<ul>" << endl;            }            cout << "<li>";            int pos = 1;            while(pos <= len)            {                if(line[pos]!=' ')                {                    break;                }                pos++;            }            cout <<dealstring(line.substr(pos,len-pos));            cout << "</li>" << endl;        }        else        {            if(star)            {                star = false;                cout << "</ul>" << endl;            }            if(!paragraph)            {                paragraph = true;                cout << "<p>";            }            else            {                cout << endl;            }            int pos = 0;            while(pos <= len)            {                if(line[pos]!=' ')                {                    break;                }                pos++;            }            cout << dealstring(line.substr(pos,len-pos));        }    }    if(star)    {        star = false;        cout << "</ul>" << endl;    }    if(paragraph)    {        paragraph = false;        cout << "</p>" << endl;    }    return 0;}/*#  Heading##  Sub-headingParagraphs are seperatedby a blank line.Text attributes _italic_.Bullet list:* apples* oranges* pearsA [link](http://example.com).*/

201612-3 权限查询

思路：这题主要就是stl的嵌套，map套map，还是思路问题，大脑一定要清晰

对于第一个特权，既然后面限定特权合法，那么完全没用啊，直接读进来就好了，getline读缓冲区，时时输出检查，这题还挺顺利，差点一发ac，就是没注释freopen有点坑了

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <stack>using namespace std;const int maxn = 1e5+10;map<string,map<string,int> > privilege;int main(){    //cout << privilege["5"] << "*"<<endl;    //freopen("in.txt","r",stdin);    ios::sync_with_stdio(false);    privilege.clear();    int p;    cin >> p;    for(int i=0;i<=p;i++)    {        string temp;        getline(cin,temp);    }    int r;    cin >> r;    for(int i=0;i<=r;i++)    {        string temp;        getline(cin,temp);        int len = temp.size();        if(len == 0)        {            continue;        }        //cout << temp << "*&^" << endl;        int pos;        string role;        temp += " ";        len++;        for(pos=0;pos<len;pos++)        {            if(temp[pos] == ' ')            {                break;            }        }        role = temp.substr(0,pos);        pos += 3;        string pri;        while(pos<len)        {            //cout << pos << "$$$" << endl;            bool nex = false;            int posstop;            for(posstop=pos;posstop<len;posstop++)            {                if(temp[posstop] == ':')                {                    break;                }                else if(temp[posstop] == ' ')                {                    nex = true;                    break;                }            }            pri = temp.substr(pos,posstop-pos);            //cout << pri << "%^&&^%" << endl;            int ran = -1;            if(nex)            {                nex = false;            }            else            {                posstop++;                ran = temp[posstop]-'0';                posstop++;            }            //cout << pri << "%^&&^%" << ran << endl;            if((privilege.find(role)!=privilege.end())&&(privilege[role].find(pri)!=privilege[role].end()))            {                if(privilege[role][pri]<ran)                {                    privilege[role][pri] = ran;                }            }            else            {                privilege[role][pri] = ran;            }            pos = posstop+1;        }    }    /*map<string,map<string,int> > :: iterator it = privilege.begin();    map<string,int> :: iterator itt;    for(;it!=privilege.end();it++)    {        cout << (it->first) <<"*"<< endl;        for(itt=(it->second).begin();itt!=(it->second).end();itt++)        {            cout << (itt->first) << "#" << (itt->second) << endl;        }    }*/    int u;    cin >> u;    for(int i=0;i<=u;i++)    {        string temp;        getline(cin,temp);        int len = temp.size();        if(len == 0)        {            continue;        }        string person;        int pos;        temp += " ";        len++;        for(pos=0;pos<len;pos++)        {            if(temp[pos] == ' ')            {                break;            }        }        person = temp.substr(0,pos);        pos += 3;        string rr;        while(pos<len)        {            int posstop;            for(posstop=pos;posstop<len;posstop++)            {                if(temp[posstop] == ' ')                {                    break;                }            }            rr = temp.substr(pos,posstop-pos);            //cout << pri << "%^&&^%" << endl;            map<string,int> :: iterator pit = privilege[rr].begin();            for(;pit!=privilege[rr].end();pit++)            {                if((privilege.find(person)!=privilege.end())&&(privilege[person].find(pit->first)!=privilege[person].end()))                {                    if(privilege[person][pit->first]<(pit->second))                    {                        privilege[person][pit->first]=(pit->second);                    }                }                else                {                    privilege[person][pit->first]=(pit->second);                }            }            pos = posstop+1;        }    }    /*map<string,map<string,int> > :: iterator it = privilege.begin();    map<string,int> :: iterator itt;    for(;it!=privilege.end();it++)    {        cout << (it->first) <<"*"<< endl;        for(itt=(it->second).begin();itt!=(it->second).end();itt++)        {            cout << (itt->first) << "#" << (itt->second) << endl;        }    }*/    int q;    cin >> q;    for(int i=0;i<=q;i++)    {        string temp;        getline(cin,temp);        int len = temp.size();        if(len == 0)        {            continue;        }        int pos;        temp += " ";        len++;        string sperson;        for(pos=0;pos<len;pos++)        {            if(temp[pos] == ' ')            {                break;            }        }        sperson = temp.substr(0,pos);        pos ++;        string pri;        bool nex = false;        int posstop;        for(posstop=pos;posstop<len;posstop++)        {            if(temp[posstop] == ':')            {                break;            }            else if(temp[posstop] == ' ')            {                nex = true;                break;            }        }        pri = temp.substr(pos,posstop-pos);        int ran = -1;        if(!nex)        {            posstop++;            ran = temp[posstop]-'0';        }        if(privilege.find(sperson) != privilege.end())        {            if(privilege[sperson].find(pri)!=privilege[sperson].end())            {                    if(ran==-1)                    {                        if(privilege[sperson][pri] == ran)                        {                            cout << "true" << endl;                        }                        else                        {                            cout << privilege[sperson][pri] << endl;                        }                    }                    else                    {                        if(privilege[sperson][pri]>=ran)                        {                            cout << "true" << endl;                        }                        else                        {                            cout << "false" << endl;                        }                    }            }            else            {                cout << "false" << endl;            }        }        else        {            cout << "false" << endl;        }    }    //cout << p << r << u << q << endl;    return 0;}

201604-3 路径解析

思路：这题也是满满的坑点啊

首先，对于合法路径，一定是由‘/’和名称组成，且名称中不含‘/’，那么利用压栈的思想，直接记录路径，最后自己生成标准格式，这样就免去了矫正格式中的大量细节

这题主要两个坑点就是‘.’，这个不起眼的点号，不仅仅可以表示当前路径，上一路径，也可以作为文件名，这个确实很意外，比较难考虑到，因为题目中只说了名称不为‘.’和‘..’，没说不能作为文件名的一部分

其他还有就是并不一定以‘.’开头才是相对路径，一切不以‘/’开头的路径均为相对路径，其实这个题目中说的很清楚，但很容易相当然就忽略误解了

最后就是空行了，这个一不小心就容易被忽略了，一不小心也就容易被改动，还是细心问题

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <stack>using namespace std;const int maxn = 1010;vector <string> contem;int clen;void correctaddress(string s){    stack <string> address;    while(!address.empty())    {        address.pop();    }    stack <string> prin;    while(!prin.empty())    {        prin.pop();    }    int len = s.size();    if(s[0]!='/'||s.empty())    {        for(int i=0;i<clen;i++)        {            address.push(contem[i]);        }    }    if(!s.empty())    {        s += '/';    }    string addr;    addr.clear();    //bool name = false;    for(int pos=0;pos<len;pos++)    {        if(s[pos]=='/')        {            if(!addr.empty())            {                if(addr == "..")                {                    if(!address.empty())                    {                        address.pop();                    }                }                else if(addr != ".")                {                    address.push(addr);                }                addr.clear();            }            //name = false;        }        /*else if(s[pos]=='.'&&(!name))        {            if(s[pos+1]=='/')            {                continue;            }            if(s[pos+1]=='.'&&s[pos+2]=='/')            {                if(!address.empty())                {                    address.pop();                }            }            else            {                addr += s[pos];                //name = true;            }        }*/        else        {            addr += s[pos];            //name = true;        }    }    if(!addr.empty())    {        if(addr == "..")        {            if(!address.empty())            {                address.pop();            }        }        else if(addr != ".")        {            address.push(addr);        }        addr.clear();    }    while(!address.empty())    {        string str = address.top();        prin.push(str);        address.pop();    }    if(prin.empty())    {        cout << "/" << endl;    }    else    {        while(!prin.empty())        {            string str = prin.top();            cout << "/" << str;            prin.pop();        }        cout << endl;    }    return ;}int main(){    //freopen("in.txt","r",stdin);    ios::sync_with_stdio(false);    int p;    cin >> p;    string ori;    getline(cin,ori);    getline(cin,ori);    contem.clear();    int len = ori.size();    string addr;    addr.clear();    for(int pos=0;pos<len;pos++)    {        if(ori[pos]=='/')        {            if(!addr.empty())            {                contem.push_back(addr);                clen++;                addr.clear();            }        }        else        {            addr += ori[pos];        }    }    if(!addr.empty())    {        contem.push_back(addr);        clen++;        addr.clear();    }    /*for(int i=0;i<(int)(contem.size());i++)    {        cout << contem[i] <<"*" << clen<< endl;    }*/    while(p--)    {        string now;        getline(cin,now);        correctaddress(now);    }    return 0;}

就此，近两年的第三道大模拟也全部刷完了，这周离出发还有7天，加油！

文章地址：http://blog.csdn.net/owen_q/article/details/78237560