hoj 3029 Dictionary 模拟队列

/*



题目:

    XML有缩进,并且会在之前的缩进基础上缩进几个空格,现在问缩进k个空格的标签



分析:

    模拟队列,及时更新空格即可







*/

#include <iostream>

#include <cstdio>

#include <cstring>

#include <vector>



using namespace std;



const int X = 20005;



int n,k;

char s[105];

char p[105];

vector<int> vec[X];



struct node

{

    char s[105];

    int num;

}q[X],ans[X];



int top,ret;



void check()

{

    if(s[0]!='<')

        return;

    if(s[1]=='/')

    {

        top--;

        return;

    }

    else

    {

        int temp = 0;

        int cnt = 0;

        bool ok = false;;

        for(int i=1;s[i];i++)

        {

            if(ok)

            {

                if(s[i]=='>')

                    break;

                temp = temp*10+s[i]-'0';

            }

            else if(s[i]!=',')

                p[cnt++] = s[i];

            if(s[i]==',')

            {

                p[cnt] = '\0';

                ok = true;

            }

        }

        if(top)

            q[top].num = q[top-1].num+temp;

        else

            q[top].num = temp;

        strcpy(q[top].s,p);

        strcpy(ans[ret].s,p);

        ans[ret].num = q[top].num;

        vec[ans[ret].num].push_back(ret);

        ret++;

        top++;

    }

}



int main()

{

    freopen("sum.in","r",stdin);

    while(scanf("%d%d ",&n,&k)!=EOF)

    {

        for(int i=0;i<n;i++)

            vec[i].clear();

        top = 0;

        ret = 0;

        for(int i=0;i<n;i++)

        {

            gets(s);

            check();

        }

        int len = vec[k].size();

        int x;

        cout<<len<<endl;

        for(int i=0;i<len;i++)

        {

            x = vec[k][i];

            printf("%s\n",ans[x].s);

        }

    }

    return 0;

}

你可能感兴趣的:(IO)