題干：

The 44th World Finals of the International Collegiate Programming Contest (ICPC 2020) will be held in Moscow, Russia. To celebrate this annual event for the best competitive programmers around the world, it is decided to host a welcome party for all??participants of the World Finals, numbered from??to??for convenience.

The party will be held in a large hall. For security reasons, all participants must present their badge to the staff and pass a security check in order to be admitted into the hall. Due to the lack of equipment to perform the security check, it is decided to open only one entrance to the hall, and therefore only one person can enter the hall at a time.

Some participants are friends with each other. There are??pairs of mutual friendship relations. Needless to say, parties are more fun with friends. When a participant enters the hall, if he or she finds that none of his or her friends is in the hall, then that participant will be unhappy, even if his or her friends will be in the hall later. So, one big problem for the organizer is the order according to which participants enter the hall, as this will determine the number of unhappy participants. You are asked to find an order that minimizes the number of unhappy participants. Because participants with smaller numbers are more important (for example the ICPC director may get the number 1), if there are multiple such orders, you need to find the lexicographically smallest one, so that important participants enter the hall first.

Please note that if participant??and??are friends, and if participant??and??are friends, it's NOT necessary that participant??and??are friends.

Input

There are multiple test cases. The first line of the input contains a positive integer?, indicating the number of cases. For each test case:

The first line contains two integers??and??(), the number of participants and the number of friendship relations.

The following??lines each contains two integers??and??(), indicating that the?-th and the?-th participant are friends. Each friendship pair is only described once in the input.

It is guaranteed that neither the sum of??nor the sum of??of all cases will exceed?.

Output

For each case, print a single integer on the first line, indicating the minimum number of unhappy participants. On the second line, print a permutation of??to?separated by a space, indicating the lexicographically smallest ordering of participants entering the hall that achieves this minimum number.

Consider two orderings??and?, we say??is lexicographically smaller than?, if there exists an integer??(), such that??holds for all?, and?.

Please, DO NOT output extra spaces at the end of each line, or your solution may be considered incorrect!

Sample Input

2 4 3 1 2 1 3 1 4 4 2 1 2 3 4

Sample Output

1 1 2 3 4 2 1 2 3 4

題目大意：有n個(gè)人m個(gè)關(guān)系，每一對(duì)關(guān)系代表這兩人認(rèn)識(shí)，關(guān)系不具有傳遞性，現(xiàn)在你要安排這n個(gè)人進(jìn)入一個(gè)舞會(huì)，如果第i個(gè)人進(jìn)入了舞會(huì)大廳但是發(fā)現(xiàn)沒(méi)有一個(gè)認(rèn)識(shí)的人，那就會(huì)不高興。問(wèn)你怎么安排順序這n個(gè)人進(jìn)入舞會(huì)大廳，使得不開(kāi)心的人數(shù)最少。如果有多種方案，輸出字典序最小的方案。

解題報(bào)告：

? 首先每一個(gè)連通塊肯定有一個(gè)人是不開(kāi)心的，并且可以做到只有一個(gè)人是不開(kāi)心的，所以有多少連通塊，就有多少人不開(kāi)心。對(duì)于字典序，只需要搞個(gè)優(yōu)先隊(duì)列遍歷就可以了。然后因?yàn)橛卸鄠€(gè)連通塊，肯定不能一個(gè)塊一個(gè)塊的遍歷，所以需要弄一個(gè)超級(jí)源點(diǎn)連到所有的起點(diǎn)上，然后再遍歷，注意連到每個(gè)聯(lián)通塊的起點(diǎn)，這個(gè)起點(diǎn)需要是整個(gè)塊的最小值，這樣肯定使得結(jié)果最優(yōu)。而讓塊的祖先節(jié)點(diǎn)是最小值，這一點(diǎn)可以用并查集稍作修改實(shí)現(xiàn)。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e6 + 5; int f[MAX],n,m; int getf(int v) {return f[v] == v ? v : f[v] = getf(f[v]); } void merge(int u,int v) {int t1 = getf(u);int t2 = getf(v);if(t1>t2) f[t1] = t2;else f[t2]=t1; } int ans[MAX],tot,vis[MAX]; vector<int> vv[MAX]; void bfs(int st) {priority_queue<int,vector<int>,greater<int> > pq;pq.push(st);while(!pq.empty()) {int cur = pq.top();pq.pop();if(vis[cur] == 1) continue;vis[cur] = 1;ans[++tot] = cur;int up = vv[cur].size();for(int i = 0; i<up; i++) {int v = vv[cur][i];if(vis[v]) continue;pq.push(v);}} } int main() {int t;cin>>t;while(t--) {tot=0;scanf("%d%d",&n,&m);for(int i = 0; i<=n; i++) f[i] = i,vv[i].clear(),vis[i]=0;for(int u,v,i = 1; i<=m; i++) {scanf("%d%d",&u,&v);vv[u].pb(v);vv[v].pb(u);merge(u,v);}int ans1 = 0;for(int i = 1; i<=n; i++) {if(f[i] == i) vv[0].pb(i),ans1++;}bfs(0);printf("%d\n",ans1);for(int i = 2; i<=tot; i++) {printf("%d%c",ans[i],i==tot ? '\n': ' ');}}return 0 ; }

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