Kth Minimum Clique_2019牛客暑期多校训练营（第二场）

题目连接：

https://ac.nowcoder.com/acm/contest/882/D

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.

Input

The first line of input contains two space-separated integers N, K.
The second line of input contains N space-separated integers wi representing the weight of each vertex.
Following N lines each contains N characters eij. There's an edge between vertex i and vertex j if and only if eij = "1".
1 <= N <= 100
1 <= K <= 10^6
0 <= wi <= 10^9

Output

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

Sample Input

2 3
1 2
01
10

Sample Output

2

Hint

题意

给出一个图的邻接矩阵，求该图的所有完全子图(点属于该图且任意两点有边)中权值第K小的价值，图权值为图中所有点的权值之和

题解：

用类似bfs的策略从0个点开始搜索，利用优先队列每次取出价值最小的子图，第K次出队的子图就是第K小的

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <bitset>

using namespace std;
typedef long long ll;
const int mx = 105;
ll v[mx];
int n, k;
int mp[mx][mx];
bitset<mx> sum[mx];

struct Node {
    bitset<mx> state;
    ll value;
    int last;
    bool operator < (Node other) const {
        return value > other.value;
    }
};


ll solve() {
    priority_queue <Node> Q;
    Node st;
    st.state.reset();
    st.value = 0;
    st.last = 0;
    Q.push(st);
    while (!Q.empty()) {
        k--;
        Node now = Q.top();
        Node next = now;
        Q.pop();
        if (k == 0) return now.value;
        for (int i = now.last+1; i <= n; i++) {
            if ((now.state & sum[i]) == now.state) {
                next = now;
                next.state[i] = 1;
                next.value += v[i];
                next.last = i;
                Q.push(next);
            }
        }
    }
    return -1;
}

char str[mx];

int main() {
    scanf("%d%d", &n, &k);

    for (int i = 1; i <= n; i++) scanf("%lld", &v[i]);
    for (int i = 1; i <= n; i++) {
        scanf("%s", str+1);
        for (int j = 1; j <= n; j++) {
            mp[i][j] = str[j]-'0';
            if (mp[i][j]) sum[i][j] = 1;
        }
    }
    printf("%lld\n", solve());
    return 0;
}