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\n"); 100 else if(oddloop)printf("YES\n"); 101 else if(M&1){ 102 printf("%s\n",higodd>=M?"YES":"NO"); 103 } 104 else { 105 printf("%s\n",(evenloop || higeven>=M)?"YES":"NO"); 106 } 107 } 108 } 109 return 0; 110 }
