题目链接:https://vjudge.net/PRoblem/UVALive-6910
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 above Output
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 5 Sample 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 1 Sample Output
Case #1: YES NO YES Case #2: YES NO Case #3: NO YES Case #4: YES
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 解法:倒着做,砍边就变成连边了,然后dsu莽一波就好了
//LA 6910#include <bits/stdc++.h>using namespace std;const int maxn = 1e5+7;int n, q, ans[maxn], a[maxn], b[maxn], fa[maxn], cnt[maxn];namespace dsu{ int par[maxn]; inline void init(){for(int i = 1; i <= n; i++) par[i] = i;} inline int find_set(int x){if(x == par[x]) return x; return par[x] = find_set(par[x]);} inline void union_set(int x, int y){x = find_set(x), y = find_set(y); if(x != y) par[x] = y;}}using namespace dsu;int main(){ int T, ks = 0; scanf("%d", &T); while(T--){ scanf("%d%d", &n, &q); for(int i = 1; i <= n; i++) scanf("%d", &fa[i]); init(); memset(cnt, 0, sizeof(cnt)); for(int i = 1; i <= q; i++){ char cmd[5]; b[i] = 0; scanf("%s%d", cmd, &a[i]); if(cmd[0] == 'Q') scanf("%d", &b[i]); else ++cnt[a[i]]; } for(int i = 1; i <= n; i++){ if(!fa[i] || cnt[i]) continue; union_set(i, fa[i]); } for(int i = q; i >= 1; i--){ if(b[i]) ans[i] = find_set(a[i]) == find_set(b[i]); else if(--cnt[a[i]] == 0 && fa[a[i]]){ union_set(a[i], fa[a[i]]); } } printf("Case #%d:/n", ++ks); for(int i = 1; i <= q; i++){ if(b[i]){ puts(ans[i] ? "YES" : "NO"); } } } return 0;}新闻热点
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