CSUOJ 1601 War (offline query and query the number of connected blocks)

CSUOJ 1601 War (offline query and query the number of connected blocks)

1601: WarTime Limit: 1 Sec Memory Limit: 128 MB
Submit: 130 Solved: 38
[Submit] [Status] [Web Board] Description

AME decided to destroy CH's country. in CH 'country, There are N ages, which are numbered from 1 to N. we say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected. to defend the country from the attack of AME, CH has decided to build some roads between some versions. let us say that two versions ages belong to the same garrison area if they are connected.
Now AME has already worked out the overall plan including which road and in which order wocould be attacked and destroyed. CH wants to know the number of garrison areas in his country after each of AME's attack.

Input

The first line contains two integers N and M-the number of ages and roads, (2 ≤ N ≤ 100000; 1 ≤ M ≤ 100000 ). each of the next M lines contains two different integers u, v (1 <= u, v <= N)-which means there is a road between u and v. the next line contains an integer Q which denotes the quantity of roads AME wants to destroy (1 ≤ Q ≤ M ). the last line contains a series of numbers each of which denoting a road as its order of appearance-different integers separated by spaces.

Output

Output Q integers-the number of garrison areas in CH's country after each of AME's attack. Each pair of numbers are separated by a single space.

Sample Input

3 11 2114 41 22 31 33 432 4 3

Sample Output

31 2 3

 

Question link: http://acm.csu.edu.cn/OnlineJudge/problem.php? Id = 1601

N vertices, m edges (side 1, side 2... edge m), q edge to be destroyed, calculate the number of connected blocks in the graph after each edge is destroyed in order

 

Question Analysis: offline query and set, reverse edge adding, calculate the number of connected blocks after all the q-entries are destroyed, and then add an edge, if two points are not in the same connected block, they are merged and the number of connected blocks is reduced by 1.

 

 

#include 
 
  #include 
  
   int const MAX = 1e5 + 5;int fa[MAX], n, m;int x[MAX], y[MAX], r[MAX];int ans, res[MAX];bool vis[MAX];void UF_set(){for(int i = 0; i <= n; i++)fa[i] = i;}int Find(int x){return x == fa[x] ? x : fa[x] = Find(fa[x]);}void Union(int a, int b){int r1 = Find(a);int r2 = Find(b);if(r1 != r2){fa[r2] = r1;ans --;}}int main(){while(scanf("%d %d", &n, &m) != EOF){UF_set();ans = n;memset(vis, false, sizeof(vis));for(int i = 1; i <= m; i++)scanf("%d %d", &x[i], &y[i]);int q;scanf("%d", &q);for(int i = 1; i <= q; i++){scanf("%d", &r[i]);vis[r[i]] = true;}for(int i = 1; i <= m; i++)if(!vis[i])Union(x[i], y[i]);for(int i = q; i >= 1; i--){res[i] = ans;Union(x[r[i]], y[r[i]]);}for(int i = 1; i < q; i++)printf("%d ", res[i]);printf("%d\n", res[q]);}}