2019牛客多校第二场 Kth Minimum Clique

题目描述

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.

输入描述:

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

输出描述:

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

示例1

输入

复制

2 3
1 2
01
10

输出

复制

2

说明

An empty set is considered as a clique.

题意：

　　找到第k小权值的团。

分析：

　　首先空集一定是最小团。

　　在空集的加入一个点，可以得到n个新集合。再逐渐往n个集合中加点，又可以得到新集合。

　　那么我们可以在加点时，保证集合的权值是逐渐变大的。

　　每次我们可以从所有的现有集合中权值最小的开始更新，所以用到优先队列。

　　按照这样逐渐加点就可以得到第k小的团，但是可能会有重复或者遗漏，从当前集合标号最大的点之后加点就不会重复遗漏了。

　　优先队列中存的就是完全子图，完全子图可以用bitset储存。

　　

///  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
",get_ans());
    return 0;
}