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=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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,);mem(in,);
         for(i=;i<=n;i++){
             scanf("%d",&a);
             to[i]=a;in[a]=;
         }
         for(i=;i<=n;i++){
             scanf("%d",&w[i]);
         }
         mem(vis,);
         higodd=higeven=oddloop=evenloop=;
         for(i=;i<=n;i++){
             if(in[i])continue;
             for(j=i,t=ok=;j && !vis[j];j=to[j]){
                 vis[j]=;
                 if(w[j]==)ok=;
                 t+=w[j];
             }
             if(vis[j])ok?oddloop=:evenloop=;
             else ok?higodd=Max(higodd,t):higeven=Max(higeven,t);
         }
         for(i=;i<=n;i++){
             if(vis[i])continue;
             for(j=i,t=ok=;!vis[j];j=to[j]){
                 vis[j]=;
                 if(w[j]==)ok=;
                 t+=w[j];
             }
             ok?oddloop=:evenloop=;
         }

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