USACO Section 4.3 Street Race - 简单搜索

    50个点...100条边....数据量很小...第一问就直接枚举去掉每个点看能否到达终点~~~N次BFS就ok了...

    第二问~~首先第二问的点肯定是第一问的点的子集~~这个应该好想到..然后就按他的描述...枚举第一问得到的点..从起点开始BFS到枚举的这个点记录能到达的点..从枚举的这个点开始BFS到终点并记录所能到达的点...若两者标记的点中没有重复则说明这个第一问的点同时满足第二问的要求~

Program:

/*  ID: zzyzzy12   LANG: C++   TASK: race3*/      #include<iostream>      #include<istream>  #include<stdio.h>     #include<string.h>      #include<math.h>      #include<stack>#include<map>#include<algorithm>      #include<queue>   #define oo 1000000000  #define ll long long  #define pi (atan(2)+atan(0.5))*2 using namespace std;  struct node{      int x,y,next;    }line[305];int n,m,_link[55],Ans1[55],Ans1Num;bool used[55];queue<int> myqueue;void getanswer1(){      int p=1,h,k;       Ans1Num=0;      for (p=1;p<n;p++)      {            while (!myqueue.empty()) myqueue.pop();            memset(used,false,sizeof(used));            used[p]=true; used[0]=true;            myqueue.push(0);            while (!myqueue.empty())            {                    h=myqueue.front();                   myqueue.pop();                   k=_link[h];                   while (k)                   {                         if (!used[line[k].y])                         {                               used[line[k].y]=true;                               myqueue.push(line[k].y);                                              }                         k=line[k].next;                         }            }            if (!used[n])                Ans1[++Ans1Num]=p;                              }      printf("%d",Ans1Num);      for (p=1;p<=Ans1Num;p++) printf(" %d",Ans1[p]);      printf("\n");      return;}void getanswer2(){      int Ans2Num,Ans2[55],p,h,k;      bool f[55];      Ans2Num=0;      for (p=1;p<=Ans1Num;p++)      {             while (!myqueue.empty()) myqueue.pop();             memset(f,false,sizeof(f));             f[Ans1[p]]=true; myqueue.push(Ans1[p]);             while (!myqueue.empty())             {                     h=myqueue.front();                     myqueue.pop();                      k=_link[h];                     while (k)                     {                           if (!f[line[k].y])                           {                                f[line[k].y]=true;                                myqueue.push(line[k].y);                                                }                           k=line[k].next;                     }             }              memset(used,false,sizeof(used));             used[0]=true; myqueue.push(0);             while (!myqueue.empty())             {                    h=myqueue.front();                    myqueue.pop();                    if (h==Ans1[p]) continue;                    if (f[h]) goto A;                     k=_link[h];                    while (k)                    {                           if (!used[line[k].y])                           {                                 used[line[k].y]=true;                                 myqueue.push(line[k].y);                                                }                           k=line[k].next;                    }                                 }             Ans2[++Ans2Num]=Ans1[p];             A: ;      }      printf("%d",Ans2Num);      for (k=1;k<=Ans2Num;k++) printf(" %d",Ans2[k]);      printf("\n");      return;}int main()  {        freopen("race3.in","r",stdin);         freopen("race3.out","w",stdout);         int i,y;       m=0; n=-1;      memset(_link,0,sizeof(_link));      while (1)      {              scanf("%d",&y);             if (y==-1) break;             n++;             while (y!=-2)             {                   m++;                   line[m].x=n; line[m].y=y;                   line[m].next=_link[n]; _link[n]=m;                   scanf("%d",&y);                  }      }       getanswer1();          getanswer2();        return 0;     }