HDU 4696 Answers 水题

http://acm.hdu.edu.cn/showproblem.php?pid=4696

由题意可知 1<=Ci<=2 而且图上一定有环

那么我们可以得出:

只要存在奇环(即Ci=1)..那么对于所有自然数Mi都能得到

否则就是偶环 只有Mi是偶数才能得到

/********************* Template ************************/
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;

#define EPS         1e-8
#define MAXN        100005
#define MOD         (int)1e9+7
#define PI          acos(-1.0)
#define DINF        (1e10)
#define LINF        ((1LL)<<50)
#define INF         (0x3f3f3f3f)
#define max(a,b)    ((a) > (b) ? (a) : (b))
#define min(a,b)    ((a) < (b) ? (a) : (b))
#define max3(a,b,c) (max(max(a,b),c))
#define min3(a,b,c) (min(min(a,b),c))
#define BUG         cout<<"BUG! "<<endl
#define line        cout<<"--------------"<<endl
#define L(t)        (t << 1)
#define R(t)        (t << 1 | 1)
#define Mid(a,b)    ((a + b) >> 1)
#define lowbit(a)   (a & -a)
#define FIN         freopen("in.txt","r",stdin)
#define FOUT        freopen("out.txt","w",stdout)
#pragma comment     (linker,"/STACK:102400000,102400000")

// typedef long long LL;
// typedef unsigned long long ULL;
// typedef __int64 LL;
// typedef unisigned __int64 ULL;
// LL gcd(LL a,LL b){ return b?gcd(b,a%b):a; }
// LL lcm(LL a,LL b){ return a*b/gcd(a,b); }

/*********************   F   ************************/
int to[MAXN];
bool vis[MAXN];
int val[MAXN];
int main()
{
    //FIN;
    //FOUT;
    int n,Q,c;
    while(~scanf("%d%d",&n,&Q)){
        memset(vis,false,sizeof(vis));
        for(int i = 1 ; i <= n ; i++)
            scanf("%d",&to[i]);
        for(int i = 1 ; i <= n ; i++)
            scanf("%d",&val[i]);
        bool odd_cir = false;
        for(int i = 1 ; i <= n && !vis[i] ; i++){
            if(val[i] == 1){
                int res = 0;
                for(int j = i ;; j = to[j]){
                    res += val[i];
                    vis[j] = true;
                    if(vis[to[j]] == true) {
                        odd_cir = true;
                        break;
                    }
                }
                if(odd_cir == true) break;
            }
        }
        for(int i = 0 ; i < Q ; i++){
            scanf("%d",&c);
            if(c <= 0) puts("NO");
            else {
                if(odd_cir == true)
                    puts("YES");
                else{
                    if(c % 2 == 0) puts("YES");
                    else puts("NO");
                }
            }
        }
    }
    return 0;
}