The 36th ACM/ICPC Asia Regional Beijing Site Online Contest - B_Eliminate Witches!

这道题很水...人名顺序就是输入字符串从左到右的顺序...边的话用一个栈来维护就行了...读入名字的时候就并将这个点与栈顶做一条边 ( 根的时候特判 )然后将这个点压入栈.. "," 与 " ) " 时做边并弹一次栈...." ( "的时候不要做处理.....最后将依次得到的边依次输出即可...开始TLE是因为可耻的用string来存名字...cout来输出...TLE得果断...然后在cout改成printf是有个地方忘记改了...又可耻的WA了一次...囧..

Program:

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int T,i,l,n,m,z[50001],p;
char s[1000010];
struct pp
{
    int x,y;
}line[100001];
char name[50010][12]; 
int main()
{ 
     scanf("%d",&T);
     getchar();
     while (T--)
     {
         memset(name,0,sizeof(name));
           gets(s);
           l=strlen(s);
           n=0; m=0;
           i=0; p=0;
           int cnt = 0;
           while (i<l)
           {
               if (s[i]>='a' && s[i]<='z')
               {
                   n++;
                   strcpy(name[n],"");
                   cnt = 0;
                   while (s[i]>='a' && s[i]<='z')
                   {
                      name[n][cnt]+=s[i];
                      i++;
                      cnt++;
                   }
                   name[n][cnt] = 0;
                   if (p)
                   {
                       m++;
                       line[m].x=z[p];
                       line[m].y=n;
                   }
                   p++;
                   z[p]=n;
               }else
               if (s[i]=='(') i++;
               else
               if (s[i]==')' || s[i]==',')
               {
                    m++;
                    line[m].x=z[p];
                    line[m].y=z[p-1];
                    p--;
                    i++;
               } 
           }
           printf("%d\n",n);
           for (i=1;i<=n;i++) printf("%s\n",name[i]);
           for (i=1;i<=m;i++) printf("%d %d\n",line[i].x,line[i].y);
           printf("\n");
     }
     return 0;
}