第k小团（Bitset+bfs）牛客第二场 -- Kth Minimum Clique

题意：

给你n个点的权值和连边的信息，问你第k小团的值是多少。

思路：

用bitset存信息，暴力跑一下就行了，因为满足树形结构，所以bfs+优先队列就ok了，其中记录下最后进入的点（以免重复跑）。

  1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
  2 #include //sprintf islower isupper
  3 #include //malloc  exit strcat itoa system("cls")
  4 #include //pair
  5 #include 
  6 #include 
  7 //#include   8 //#include
  9 #include 
 10 #include 
 11 #include <set>
 12 #include <string.h>//strstr substr
 13 #include <string>
 14 #include //srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 15 #include 
 16 #include 
 17 #include //priority_queue, greater > q;//less
 18 #include //emplace_back
 19 //#include 
 20 //#include //reverse(a,a+len);// ~ ! ~ ! floor
 21 #include //sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 22 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 23 #define mem(a,b) memset(a,b,sizeof(a))
 24 #define fo(a,b,c) for(a=b;a<=c;++a)//register int i
 25 #define fr(a,b,c) for(a=b;a>=c;--a)
 26 #define pr printf
 27 #define sc scanf
 28 void swapp(int &a,int &b);
 29 double fabss(double a);
 30 int maxx(int a,int b);
 31 int minn(int a,int b);
 32 int Del_bit_1(int n);
 33 int lowbit(int n);
 34 int abss(int a);
 35 //const long long INF=(1LL<<60);
 36 const double E=2.718281828;
 37 const double PI=acos(-1.0);
 38 const int inf=(1<<29);
 39 const double ESP=1e-9;
 40 const int mod=(int)1e9+7;
 41 const int N=(int)1e6+6;
 42  
 43 int val[105];
 44 struct node
 45 {
 46     int End;
 47     long long V;
 48     bitset<105> clique;//团
 49     friend bool operator<(node a,node b)
 50     {
 51         return a.V>b.V;
 52     }
 53 };
 54 bitset<105> mp[105];
 55 priority_queue q;
 56  
 57 int main()
 58 {
 59     int n,k,i,j;
 60     sc("%d%d",&n,&k);
 61     for(i=1;i<=n;++i)
 62         sc("%d",&val[i]);
 63     for(i=1;i<=n;++i)
 64         for(j=1;j<=n;++j)
 65         {
 66             int t;
 67             sc("%1d",&t);
 68             mp[i][j]=t;
 69         }
 70     bitset<105> temp;
 71     q.push({0,0,temp});
 72     while(!q.empty())
 73     {
 74         node now=q.top();q.pop();
 75         k--;
 76         if(k==0)
 77         {
 78             pr("%lld\n",now.V);
 79             return 0;
 80         }
 81         for(i=now.End+1;i<=n;++i)
 82         {
 83             if((now.clique&mp[i])==now.clique)
 84             {
 85                 node v=now;
 86                 v.clique[i]=1;
 87                 v.End=i;
 88                 v.V+=val[i];
 89                 q.push(v);
 90             }
 91         }
 92     }
 93     pr("-1\n");
 94     return 0;
 95 }
 96  
 97 /**************************************************************************************/
 98  
 99 int maxx(int a,int b)
100 {
101     return a>b?a:b;
102 }
103  
104 void swapp(int &a,int &b)
105 {
106     a^=b^=a^=b;
107 }
108  
109 int lowbit(int n)
110 {
111     return n&(-n);
112 }
113  
114 int Del_bit_1(int n)
115 {
116     return n&(n-1);
117 }
118  
119 int abss(int a)
120 {
121     return a>0?a:-a;
122 }
123  
124 double fabss(double a)
125 {
126     return a>0?a:-a;
127 }
128  
129 int minn(int a,int b)
130 {
131     return aa:b;
132 }

 

转载于:https://www.cnblogs.com/--HPY-7m/p/11436879.html