UVALive 6910 Cutting Tree 并查集

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 ofN 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 thisquery.

• 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 beginswith two integers: N and K (1 ≤ N ≤ 20, 000; 1 ≤ K ≤ 5, 000) denoting the number of nodes in theforest and the number of queries respectively. The nodes are numbered from 1 to N. The next linecontains N integers Pi (0 ≤ Pi ≤ N) denoting the parent of i-th node respectively. Pi = 0 means thatnode i does not have any parent (i.e. it’s a root of a tree). You are guaranteed that the given inputcorresponds to a valid forest. The next K lines represent the queries. Each query is in the form of ‘Cx’ 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 ‘Qa b’ query in the input, output either ‘YES’ or ‘NO’ (without quotes) in a line whether there is a pathfrom 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.Figure 3. Figure 4. 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

Source

UESTC 2016 Summer Training #19

UVALive 6910

My Solution

简单并查集

给出一片森林， 然后执行

1）切断 x和x的父节点的边，    //  查询的时候不进行路径压缩， 然后直接 father[ x] = x

2）查询 x， y 时候是相连通的 //分别找x y的根， _find(x) == _find(y)

复杂度 O(n)

#include <iostream>#include <cstdio>#include <cstring>using namespace std;typedef long long LL;const int maxn = 2*1e5 + 8;int father[maxn], _rank[maxn], sz[maxn], trees[maxn];bool flag;inline void DisjointSet(int n){    for(int i = 0; i <= n; i++){        father[i] = i;        trees[i] = 1;    }}int x, y;int _find(int v, int u)  //请忽略这个 _find(){    if(father[v] == u) return -1;    return father[v] == v ? v : _find(father[v], u);}int _find(int v){    //if(father[v] == u) return -1;    return father[v] == v ? v : _find(father[v]);}int main(){    #ifdef LOCAL    freopen("a.txt", "r", stdin);    //freopen("b.txt", "w", stdout);    #endif // LOCAL    int T, n, k, val, u, kase = 0;    char op;    scanf("%d", &T);    while(T--){        scanf("%d%d", &n, &k);        for(int i = 1; i <= n; i++){            scanf("%d", &val);            if(val != 0) father[i] = val;            else father[i] = i;        }        printf("Case #%d:\n", ++kase);        while(k--){            getchar(); //!!!!!!            scanf("%c", &op);            if(op == 'C'){                scanf("%d", &u);                father[u] = u;            }            else{                scanf("%d%d", &x, &y);                if(_find(x) == _find(y)) printf("YES\n");                else printf("NO\n");            }        }    }    return 0;}

  Thank you!

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