Title:
Give you n points, m edges, no double edges, no self-loops
If a sequence is satisfied, all are in 1-N and there is an edge connection between the two points.
There is also 1 0 indicating whether the number of times the path passes through point i is odd or even
answer:
1. How to express it?
(1) Use a vector to represent the input, and then use auto to get the corresponding value.
(2) dis[i][j] indicates that the current state is i "represented in binary" and the number is j.
2. How to do it?
(1) Why do you want u-1, v-1 in the first place? Because you want to use binary operations, such as 0001, it is (1<<0) instead of (1<<1)
(2) BFS, start from a number and search backwards, if you haven't searched, continue to search, then +1, if you have searched, then don't
//
#include <iostream>
#include <vector>
#include <queue>
#define int long long
using namespace std;
const int N = (1 << 17);
const int INF = 0x3f3f3f3f3f3f3f3f;
int dis[N][17];
vector<int> G[20];
queue<pair<int, int>> Q;
int n,m;
signed main() {
cin>>n>>m;
for(int i=1;i<=m;i++){
int u,v;
cin>>u>>v;
u--,v--;
G[u].push_back(v);
G[v].push_back(u);
}
int M = 1<<n;
for(int i=0;i<M;i++)
for(int j=0;j<n;j++)
dis[i][j]=INF;
// 这里为啥要一开始初始化为1呢？
// 因为我一个数，length为1！
for(int i=0;i<n;i++){
dis[1<<i][i]=1;
Q.push({1<<i,i});
}
// BFS
while(Q.size()){
int s = Q.front().first;
int v = Q.front().second;
Q.pop();
for(auto u : G[v]){
int ns = s ^ (1<<u);
if(dis[ns][u]<INF) continue;
dis[ns][u] = dis[s][v] + 1;
// 这就是挑出来的那一个
Q.push({ns,u});
}
}
int ans = 0;
// 最后是对所有情况进行加总
for(int i=1;i<M;i++){
int mm = INF;
for(int j=0;j<n;j++){
mm = min(mm,dis[i][j]);
}
ans += mm;
}
cout<<ans<<endl;
}