HDU-4696 Answers 纯YY

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

　　题意：给一个图，每个点的出度为1，每个点的权值为1或者2。给n个询问，问是否能找到一条路径的权值和M。。。

　　首先由于每个点的出度为1，所以必然存在环。容易证明，一个环中存在1或者与环相连的路径存在权值为1的节点，那么必然每个数都能组成，如果没有1那么所有偶数都能组成。。。

　　可以无视代码><

//STATUS:C++_AC_125MS_1988KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
//#include <map>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=100010;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int to[N],in[N],vis[N],w[N];
int n,m;

int main() {
 //   freopen("in.txt", "r", stdin);
    int i,j,a,t,ok,M;
    int higodd,higeven,oddloop,evenloop;
    while(~scanf("%d%d",&n,&m))
    {
        mem(to,0);mem(in,0);
        for(i=1;i<=n;i++){
            scanf("%d",&a);
            to[i]=a;in[a]=1;
        }
        for(i=1;i<=n;i++){
            scanf("%d",&w[i]);
        }
        mem(vis,0);
        higodd=higeven=oddloop=evenloop=0;
        for(i=1;i<=n;i++){
            if(in[i])continue;
            for(j=i,t=ok=0;j && !vis[j];j=to[j]){
                vis[j]=1;
                if(w[j]==1)ok=1;
                t+=w[j];
            }
            if(vis[j])ok?oddloop=1:evenloop=1;
            else ok?higodd=Max(higodd,t):higeven=Max(higeven,t);
        }
        for(i=1;i<=n;i++){
            if(vis[i])continue;
            for(j=i,t=ok=0;!vis[j];j=to[j]){
                vis[j]=1;
                if(w[j]==1)ok=1;
                t+=w[j];
            }
            ok?oddloop=1:evenloop=1;
        }

        for(i=0;i<m;i++){
            scanf("%d",&M);
            if(M<=0)printf("NO
");
            else if(oddloop)printf("YES
");
            else if(M&1){
                printf("%s
",higodd>=M?"YES":"NO");
            }
            else {
                printf("%s
",(evenloop || higeven>=M)?"YES":"NO");
            }
        }
    }
    return 0;
}