题目连接:
Description
Tree in graph theory refers to any connected graph (of nodes and edges) which has no simple cycle,
while forest corresponds to a collection of one or more trees. In this problem, you are given a forest of N nodes (of rooted trees) and K queries. Each query is in the form of: • C x : remove the edge connecting node and its parent. If node has no parent, then ignore this query. • Q a b : output ‘YES’ if there is a path from node to node in the forest; otherwise, ‘NO’. For example, let the initial forest is shown by Figure 1. Figure 1. Figure 2. Let’s consider the following queries (in order): 1) Q 5 7 : output YES. 2) C 2 : remove edge (2, 1) — the resulting forest is shown in Figure 2. 3) Q 5 7 : output NO, as there is no path from node 5 to node 7 in Figure 2. 4) Q 4 6 : output YES.Input
The first line of input contains an integer T (T ≤ 50) denoting the number of cases. Each case begins
with two integers: N and K (1 ≤ N ≤ 20, 000; 1 ≤ K ≤ 5, 000) denoting the number of nodes in the forest and the number of queries respectively. The nodes are numbered from 1 to N. The next line contains N integers Pi (0 ≤ Pi ≤ N) denoting the parent of i-th node respectively. Pi = 0 means that node i does not have any parent (i.e. it’s a root of a tree). You are guaranteed that the given input corresponds to a valid forest. The next K lines represent the queries. Each query is in the form of ‘C x’ or ‘Q a b’ (1 ≤ x, a, b ≤ N), as described in the problem statement aboveOutput
For each case, output ‘Case #X:’ in a line, where X is the case number starts from 1. For each ‘Q
a b’ query in the input, output either ‘YES’ or ‘NO’ (without quotes) in a line whether there is a path from node a to node b in the forest. Explanation for 2nd sample case: The initial forest is shown in Figure 3 below. 1) C 3 : remove edge (3, 2) — the resulting forest is shown in Figure 4. 2) Q 1 2 : output YES. 3) C 1 : remove edge (1, 2) — the resulting forest is shown in Figure 5. 4) Q 1 2 : output NO as there is no path from node 1 to node 2 in Figure 5Sample Input
4
7 4 0 1 1 2 2 2 3 Q 5 7 C 2 Q 5 7 Q 4 6 4 4 2 0 2 3 C 3 Q 1 2 C 1 Q 1 2 3 5 0 3 0 C 1 Q 1 2 C 3 C 1 Q 2 3 1 1 0 Q 1 1Sample Output
Case #1:
YES NO YES Case #2: YES NO Case #3: NO YES Case #4: YESHint
题意
给你个森林,俩操作,1是砍掉与他父亲的连边,2是查询xy是否在同一个连通块里面
题解:
倒着做,砍边就变成连边了,然后并茶几莽一波就好了
代码
#includeusing namespace std;const int maxn = 2e4+7;int cas = 0;int fa[maxn];int e[maxn];int flag[maxn];int a[maxn],b[maxn],c[maxn];;int fi(int x){ if(x==fa[x])return x; return fa[x]=fi(fa[x]);}void init(){ memset(flag,0,sizeof(flag));}void solve(){ init(); vector ans; int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) fa[i]=i; for(int i=1;i<=n;i++) scanf("%d",&e[i]); for(int i=1;i<=m;i++){ string s;cin>>s; if(s[0]=='C'){ a[i]=1; scanf("%d",&b[i]); flag[b[i]]++; }else{ a[i]=0; scanf("%d%d",&b[i],&c[i]); } } for(int i=1;i<=n;i++){ if(flag[i]==0&&e[i]!=0){ fa[fi(i)]=fi(e[i]); } } for(int i=m;i>=1;i--){ if(a[i]==1){ flag[b[i]]--; if(flag[b[i]]==0&&e[b[i]]!=0) fa[fi(b[i])]=fi(e[b[i]]); }else{ if(fi(b[i])==fi(c[i]))ans.push_back(1); else ans.push_back(0); } } for(int i=ans.size()-1;i>=0;i--){ if(ans[i])printf("YES\n"); else printf("NO\n"); }}int main(){ //freopen("1.txt","r",stdin); int t; scanf("%d",&t); while(t--){ printf("Case #%d:\n",++cas); solve(); } return 0;}