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;}