2019 cattle off more school second field Kth Minimum Clique

 

Title Description

Given a vertex-weighted graph with N vertices, find out the K-th minimum weighted clique.

A subset of vertices of an undirected graph is called clique if and only if every two distinct vertices in the subset are adjacent. The weight of a clique is the summation of the weight of vertices in it.

Enter a description:

The first line of input contains two space-separated integers N, K.
The second line of input contains N space-separated integers ${\mathrm{wiwi\; representing\; the\; weight\; of\; each\; vertex.Following\; N\; lines\; each\; contains\; N\; characters\; eijeij.\; There\text{'}s\; an\; edge\; between\; vertex\; i\; and\; vertex\; j\; if\; and\; only\; if\; eij="1"eij="1".1\le N\le 1001\le K\le 1e60\le wi\le 1e9eij\in "01"eii="0"eij=eji}}^{}$

Output Description:

Output one line containing an integer representing the answer. If there's less than K cliques, output $\text{"-1""-1".}$

Example 1

Entry

copy

2 3
1 2
01
10

Export

copy

2

Explanation

An empty set is considered as a clique.

Meaning of the questions:

　　K find small groups of weights.

analysis:

　　First, the smallest group must be empty set.

　　In a point of addition of the empty set, a new set of n can be obtained. Then gradually added to the set of n points, and can obtain a new collection.

　　Then we can add a little at the time, to ensure the right set of values ​​is progressively larger.

　　Every time we can start with minimal updates from all of the existing set of weights, so use the priority queue.

　　According to this point can be gradually added to give a small group of k, but there may be repeated or missing, then the reference current set point plus the maximum point does not duplicate the missing.

　　Stored in the priority queue is complete subgraph, complete subgraph can bitset stored.

　　

///  author:Kissheart  ///
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<vector>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<deque>
#include<ctype.h>
#include<map>
#include<bitset>
#include<set>
#include<stack>
#include<string>
#define INF 0x3f3f3f3f
#define FAST_IO ios::sync_with_stdio(false)
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MAX=1e5+10;
const int mod=1e9+7;
typedef long long ll;
using namespace std;
#define gcd(a,b) __gcd(a,b)
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
inline ll qpow(ll a,ll b){ll r=1,t=a; while(b){if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;}
inline ll inv1(ll b){return qpow(b,mod-2);}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll r=exgcd(b,a%b,y,x);y-=(a/b)*x;return r;}
inline ll read(){ll x=0,f=1;char c=getchar();for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;for(;isdigit(c);c=getchar()) x=x*10+c-'0';return x*f;}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
int n,k;
int a[105];
bitset<105>bit[105];
struct node
{
    bitset<105>now;
    ll val;
    int last;
    bool operator < (const node &a) const
    {
        return val>a.val;
    }
};

ll get_ans()
{
    priority_queue<node>q;
    node no;
    no.now.reset();
    no.val=0;
    no.last=0;
    q.push(no);

    while(!q.empty())
    {
        no=q.top();
        q.pop();

        if(--k == 0)
            return no.val;

        for(int i=no.last;i<n;i++)
        {
            if(!no.now[i] && ((no.now & bit[i])==no.now) )
            {
                no.now.set(i);
                q.push(node{no.now,no.val+a[i],i+1});
                no.now.reset(i);
            }
        }
    }
    return -1;
}
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++) scanf("%d",&a[i]);
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            int x;
            scanf("%1d",&x);
            bit[i].set(j,x);
        }
    }
    printf("%lld\n",get_ans());
    return 0;
}

View Code