【HDOJ5943】Kingdom of Obsession(数论)

题意:给定n个人,n个座位,人的编号是【1,n】,座位的编号是【s+1,s+n】,编号为i的人能坐在编号为j的座位上的条件是j%i=0

问是否存在一组方案使得座位和人一一对应

n,s<=1e9

思路:首先考虑如果区间【1,n】和【s+1,s+n】有重叠则重叠部分必然座位和人两两编号相同对应,因为这样能使解的空间更加宽松

现在考虑两端无重叠且等长的区间,显然如果靠后那段区间中出现了大于1个质数则他们都必须要用1,必然无解

根据质数密度分布可以近似算出阈值(据tls说是60?),如果区间长度大于阈值则必然无解,否则跑二分图最大匹配

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 typedef unsigned int uint;
  5 typedef unsigned long long ull;
  6 typedef long double ld;
  7 typedef pair<int,int> PII;
  8 typedef pair<ll,ll> Pll;
  9 typedef vector<int> VI;
 10 typedef vector<PII> VII;
 11 typedef pair<ll,ll>P;
 12 #define N  2000000+10
 13 #define M  200000+10
 14 #define INF 1e9
 15 #define fi first
 16 #define se second
 17 #define MP make_pair
 18 #define pb push_back
 19 #define pi acos(-1)
 20 #define mem(a,b) memset(a,b,sizeof(a))
 21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 23 #define lowbit(x) x&(-x)
 24 #define Rand (rand()*(1<<16)+rand())
 25 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 26 #define ls p<<1
 27 #define rs p<<1|1
 28 #define fors(i) for(auto i:e[x]) if(i!=p)
 29 
 30 const int MOD=998244353,inv2=(MOD+1)/2;
 31       //int p=1e4+7;
 32       //double eps=1e-6;
 33       int dx[4]={-1,1,0,0};
 34       int dy[4]={0,0,-1,1};
 35 
 36 int head[N],vet[N],nxt[N],match[N],flag[N],a[N],b[N],tot;
 37 
 38 int read()
 39 {
 40    int v=0,f=1;
 41    char c=getchar();
 42    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 43    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 44    return v*f;
 45 }
 46 
 47 ll readll()
 48 {
 49    ll v=0,f=1;
 50    char c=getchar();
 51    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 52    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 53    return v*f;
 54 }
 55 
 56 void add(int a,int b)
 57 {
 58     nxt[++tot]=head[a];
 59     vet[tot]=b;
 60     head[a]=tot;
 61 }
 62 
 63 int dfs(int u)
 64 {
 65     flag[u]=1;
 66     int e=head[u];
 67     while(e)
 68     {
 69         int v=vet[e];
 70         if(!match[v]||!flag[match[v]]&&dfs(match[v]))
 71         {
 72             match[v]=u;
 73             return 1;
 74         }
 75         e=nxt[e];
 76     }
 77     return 0;
 78 }
 79 
 80 int main()
 81 {
 82     //freopen("1.in","r",stdin);
 83     //freopen("1.out","w",stdout);
 84     int cas=read();
 85     rep(v,1,cas)
 86     {
 87         int n=read(),s=read();
 88         int x1,y1,x2,y2;
 89         if(n<s+1) x1=1,y1=n,x2=s+1,y2=s+n;
 90          else x1=1,y1=s,x2=n+1,y2=s+n;
 91         if(y1-x1>100) printf("Case #%d: No\n",v);
 92          else
 93          {
 94              int n=0;
 95              rep(i,x1,y1)
 96              {
 97                  n++; a[n]=i;
 98              }
 99              n=0;
100              rep(i,x2,y2)
101              {
102                  n++; b[n]=i;
103              }
104              rep(i,1,n) head[i]=match[i]=0;
105              tot=0;
106              rep(i,1,n)
107               rep(j,1,n)
108                if(b[j]%a[i]==0) add(i,j);
109              int ans=0;
110              rep(i,1,n)
111              {
112                  rep(j,1,n) flag[j]=0;
113                  if(dfs(i)) ans++;
114              }
115              if(ans==n) printf("Case #%d: Yes\n",v);
116               else printf("Case #%d: No\n",v);
117          }
118     }
119     return 0;
120 }

 

posted on 2019-11-25 19:07  myx12345  阅读(173)  评论(0编辑  收藏  举报

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