topic:
War to maintain connectivity between various cities is very important. This question requires you to write a program alarm, when the city lost a lead country is split into multiple zones can not communicate, it emits red alert. Note: If the country should not have full connectivity, k is divided regions, a lost city does not alter the connectivity between other cities, do not alarm.
Input formats:
Input two integers are given in the first row N(0 < N ≤ 500) and M( ≤ 5000), respectively, the number of city (city then default from 0 to N-1 numbered) and connections of the two cities via strip. Subsequently Mrows each given two cities connected via a number, separated by a space therebetween. It gives information was captured after city information, that is a positive integer Kand subsequently Ka number of cities were captured.
Note: Enter the city was captured guarantee given number is legitimate and no repeat, but does not guarantee passage given no repeat.
Output formats:
Each city was captured, if it will change the connectivity of the entire country, the output Red Alert: City k is lost!, which kis the city's number; otherwise, only the output City k is lost.can be. If the country lost its last city, then add a line output Game Over..
Sample input:
5 4
0 1
1 3
3 0
0 4
5
1 2 0 4 3
Sample output:
City 1 is lost.
City 2 is lost.
Red Alert: City 0 is lost!
City 4 is lost.
City 3 is lost.
Game Over.
①在判断是否改变连通性的时候，我第一个想法是每一个地点都找到pre，然后每次将两个地点加入并查集的时候都会不断更新pre使联通的地点的pre（即 爸爸 ）都是相同的，最后我把新的pre数组和旧的pre数组进行比较看是否除了lost的城市之外有改变。
我当时pre数组的记录方式是，每个pre记录共同的“爸爸”，而“爸爸”的pre是自己本身。
这样出现的问题是，即使没有改变连通性每次lost一个城市重新用并查集得到的“爸爸”不一定相同。
②后来看了同学的代码，发现可以用 判断联通集合的个数 即爸爸的个数来判断是否联通。于是 我根据前后两次的“爸爸”数量是否相通来判断是否联通
但是，我设置的pre[爸爸]为爸爸本身，如果是一个本来就孤立的城市，那么查询的时候就会把它当成一个爸爸。但实际这个城市的存在与否不会影响到连通性。
③再后来我再看了同学代码，将pre[爸爸]改为-1，在数爸爸个数的时候就数-1的个数（此时-1的个数不再是连通块的个数了，但是可以根据-1个数的变化来判断是否改变了连通性），这样可以彻底不考虑那些孤立的城市。
但是如果是不孤立的尾端的城市lost了，结果就会多出一个-1，但是实际上也没有改变连通性。
④然后我就用一个变量death记录当前的lost城市，然后再数-1个数的时候跳过death。
但是也是因此-1的数量没有更新，而在下一次找爸爸数量的时候又将这个-1算了进来，因此错误。
⑤最后我又又看了同学代码。因为如果是尾端的城市lost，那么-1个数就会比原来多一个；如果是影响连通性的城市lost，因为改变连通性多一个“爸爸”， 所以多了一个-1， 而lost的城市的pre也变成-1，那么最后多出来的-1就有两个。
所以最后判断是否改变联通性的方式是 -1的个数是否与原来相等或者比原来多一个，如果是那么没有改变连通性，反之改变。
代码：1 #include <iostream> 2 #include <fstream> 3 #include <cstring> 4 #include <algorithm> 5 6 using namespace std; 7 int N, M, areanum=-1; 8 const int mx=510; 9 int connection[mx][mx]; 10 int root[mx],pre[mx]; 11 12 int findroot(int vs); 13 int findset(); 14 void makefamily(int v1, int v2); 15 int main(){ 16 fill(connection[0], connection[0]+mx*mx, -1); 17 fill(pre, pre+mx, -1); 18 19 cin>>N>>M; 20 for(int i=1;i<=M;i++){ 21 int a, b; 22 cin>>a>>b; 23 connection[a][b]=1; 24 connection[b][a]=1; 25 makefamily(a, b); 26 } 27 findset(); 28 int K; 29 cin>>K; 30 int num=N; 31 for(int i=1;i<=K;i++){ 32 int death; 33 cin>>death; 34 for(int city=0;city<N;city++){ 35 if(connection[city][death]==1||connection[death][city]==1){ 36 connection[city][death]=-1; 37 connection[death][city]=-1; 38 } 39 } 40 fill(pre, pre+mx, -1); 41 42 for(int i=0;i<N;i++){ 43 for(int j=0;j<N;j++){ 44 if(connection[i][j]==1) 45 makefamily(i,j); 46 } 47 } 48 int flag=findset(); 49 50 (flag==-1)?printf("Red Alert: "):0; 51 printf("City %d is lost", death); 52 (flag==-1)?puts("!"):puts("."); 53 num--; 54 } 55 if(num==0) cout<<"Game Over."<<endl; 56 } 57 int findset(){ 58 int t=0, Fake = areanum; 59 for ( int I = 0 ; I <N; I ++ ) { 60 IF (pre [I] == - . 1 ) T ++ ; 61 is } 62 is areanum = T; 63 is return (T == Fake || Fake == + T . 1 )? . 1 : - . 1 ; // determines whether Unicom 64 65 } 66 67 void makefamily ( int V1, int V2) { 68 int T = V2; 69 V1 = the findroot (V1); 70 v2=findroot(v2); 71 if(v1!=v2){ 72 pre[v2]=v1; 73 } 74 } 75 int findroot(int vs){ 76 while(pre[vs]!=-1) vs=pre[vs]; 77 return vs; 78 }
Later Baidu a bit and found dfs can also be used to do this question hhh.
As https://blog.csdn.net/sgh666666/article/details/79752816