URAL 1242 Werewolf（DFS）

Posted 2020-08-09 贱人方

Werewolf

Time limit: 1.0 second
Memory limit: 64 MB

Knife. Moonlit night. Rotten stump with a short black-handled knife in it. Those who know will understand. Disaster in the village. Werewolf.

There are no so many residents in the village. Many of them are each other‘s relatives. Only this may help to find the werewolf. The werewolf is merciless, but his descendants never become his victims. The werewolf can drown the village in blood, but he never kills his ancestors.

It is known about all the villagers who is the child of whom. Also, the sad list of the werewolf‘s victims is known. Your program should help to determine the suspects. It would be a hard task, if a very special condition would not hold. Namely, citizens of the village are not used to leave it. If some ancestor of some citizen lives in the village, then also his immediate ancestor does. It means, that, for example, if the father of the mother of some citizen still lives in the village, than also his mother still lives.

Input

The first line contains an integer N, 1 < N ≤ 1000, which is the number of the villagers. The villagers are assigned numbers from 1 to N. Further is the description of the relation "child-parent": a sequence of lines, each of which contains two numbers separated with a space; the first number in each line is the number of a child and the second number is the number of the child‘s parent. The data is correct: for each of the residents there are no more than two parents, and there are no cycles. The list is followed by the word "BLOOD" written with capital letters in a separate line. After this word there is the list of the werewolf‘s victims, one number in each line.

Output

The output should contain the numbers of the residents who may be the werewolf. The numbers must be in the ascending order and separated with a space. If there are no suspects, the output should contain the only number 0.

Samples

input	output
8 1 3 3 6 4 5 6 2 4 6 8 1 BLOOD 3 8	4 5 7
6 1 2 3 2 1 4 3 4 2 6 5 2 5 4 BLOOD 2 5	0

Problem Author: Leonid Volkov

【分析】简单的递归。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 0x3f3f3f3f
#define mod 10000
typedef long long ll;
using namespace std;
const int N=1005;
const int M=100005;
int n,m,k=0,a,b;
int vis[N];
char str[20];
bool flag=false;
vector<int>vec,parent[N],chil[N];
void dfs(int u,int f)
{//printf("!!!%d %d\n",u,f);
    vis[u]=1;
    if(!f){
        for(int i=0;i<parent[u].size();i++){
                //printf("@@@%d\n",parent[u][i]);
            if(!vis[parent[u][i]])dfs(parent[u][i],f);
        }
    }
    else {
        for(int i=0;i<chil[u].size();i++){
            if(!vis[chil[u][i]])dfs(chil[u][i],f);
        }
    }
}
void in(char *ch){
    int fff=0,aa=0,bb=0;
    char A[20],B[20];
    for(int i=0;i<strlen(ch);i++){
        if(ch[i]==‘ ‘){fff=1;continue;}
        if(!fff)A[aa++]=ch[i];
        else B[bb++]=ch[i];
    }
    A[aa]=0;B[bb]=0;
    a=atoi(A);
    b=atoi(B);
    return;
}
int main() {
    scanf("%d",&n);getchar();

    while(gets(str)!=NULL){k++;
        if(str[0]==‘B‘){flag=true;continue;}
        in(str);
        if(!flag)parent[a].push_back(b),chil[b].push_back(a);
        if(flag)vec.push_back(a);//if(k>=9)break;
    }
    for(int i=0;i<vec.size();i++){
        int v=vec[i];
        dfs(v,0);
        dfs(v,1);
    }
    flag=false;
    for(int i=1;i<=n;i++)if(!vis[i])printf("%d ",i),flag=true;
    if(!flag)printf("0");
    printf("\n");
    return 0;
}

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