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><

  1 //STATUS:C++_AC_125MS_1988KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 //#include <map>
 23 using namespace std;
 24 #pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef __int64 LL;
 34 typedef unsigned __int64 ULL;
 35 //const
 36 const int N=100010;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=1000000007,STA=8000010;
 39 const LL LNF=1LL<<60;
 40 const double EPS=1e-8;
 41 const double OO=1e15;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 int to[N],in[N],vis[N],w[N];
 59 int n,m;
 60 
 61 int main() {
 62  //   freopen("in.txt", "r", stdin);
 63     int i,j,a,t,ok,M;
 64     int higodd,higeven,oddloop,evenloop;
 65     while(~scanf("%d%d",&n,&m))
 66     {
 67         mem(to,0);mem(in,0);
 68         for(i=1;i<=n;i++){
 69             scanf("%d",&a);
 70             to[i]=a;in[a]=1;
 71         }
 72         for(i=1;i<=n;i++){
 73             scanf("%d",&w[i]);
 74         }
 75         mem(vis,0);
 76         higodd=higeven=oddloop=evenloop=0;
 77         for(i=1;i<=n;i++){
 78             if(in[i])continue;
 79             for(j=i,t=ok=0;j && !vis[j];j=to[j]){
 80                 vis[j]=1;
 81                 if(w[j]==1)ok=1;
 82                 t+=w[j];
 83             }
 84             if(vis[j])ok?oddloop=1:evenloop=1;
 85             else ok?higodd=Max(higodd,t):higeven=Max(higeven,t);
 86         }
 87         for(i=1;i<=n;i++){
 88             if(vis[i])continue;
 89             for(j=i,t=ok=0;!vis[j];j=to[j]){
 90                 vis[j]=1;
 91                 if(w[j]==1)ok=1;
 92                 t+=w[j];
 93             }
 94             ok?oddloop=1:evenloop=1;
 95         }
 96 
 97         for(i=0;i<m;i++){
 98             scanf("%d",&M);
 99             if(M<=0)printf("NO
");
100             else if(oddloop)printf("YES
");
101             else if(M&1){
102                 printf("%s
",higodd>=M?"YES":"NO");
103             }
104             else {
105                 printf("%s
",(evenloop || higeven>=M)?"YES":"NO");
106             }
107         }
108     }
109     return 0;
110 }