LOJ #109. query the set,

LOJ #109. query the set,
Memory limit: 256 MiB time limit: 2000 ms standard input/output question type: traditional evaluation method: text comparison uploaded by: Anonymous Submission submission record statistics discussion 1 Test Data question description

This is a template question.

Maintains an undirected graph of an nnn point. supported:

• Add an undirected edge connecting uuu and vvv
• Query the connectivity between uuu and vvv

Because the data in this question is large, a special output method is used for output: 000 or 111 represents the answer to each query, and the answers to each query are arranged from left to right at a time, treat the resulting string as a binary number and output the binary number mod 998244353 \ text {mod }~ The value of 992134366mod 998244353.

Input Format

The first line contains two integers n, mn, mn, m, indicating the number of vertices and the number of operations.

Next, the mmm row contains three integers, op, u, v \ text {op}, u, vop, u, and v.

• If op = 0 \ text {op} = 0op = 0, a undirected edge connecting uuu and vvv is added;
• If op = 1 \ text {op} = 1op = 1, the connectivity between uuu and vvv is queried.

Output Format

A row contains an integer to indicate the answer.

Sample Input

3 61 1 00 0 11 0 11 1 20 2 11 2 1

Sample output

5

Example

The answer string is 101101101.

Data range and prompt

N ≤ 4000000, m ≤ 8000000n \ le 4000000, m \ le 8000000n ≤ 4000000, m ≤ 8000000

Show category labels

 

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1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 using namespace std; 6 const int MAXN = 8000001; 7 const int mod = 998244353; 8 inline void read (int & n) 9 {10 char c = '+'; bool flag = 0; n = 0; 11 while (c <'0' | c> '9') c = '-'? Flag = 1, c = getchar (): c = getchar (); 12 while (c> = '0' & c <= '9 ') n = n * 10 + c-48, c = getchar (); 13} 14 int fa [MAXN]; 15 int size [MAXN]; 16 int n, m; 17 string p; 18 int find (int x) 19 {return fa [x] = x? Fa [x]: fa [x] = find (fa [x]);} 20 int query (int x, int y) 21 {return find (x) = find (y);} 22 void unionn (int x, int y) 23 {24 int fx = find (x); int fy = find (y ); 25 if (fx! = Fy) 26 {27 if (size [fx]> size [fy]) swap (fx, fy); 28 fa [fx] = fy; size [fy] + = size [fx]; 29 // fa [fx] = fy; 30} 31} 32 int ans = 0; 33 int main () 34 {35 // freopen (". in "," r ", stdin); 36 // freopen (". out "," w ", stdout); 37 read (n); read (m); 38 for (int I = 1; I <= n; I ++) fa [I] = I; 39 for (int I = 1; I <= m; I ++) 40 {41 int how; read (how); 42 if (how) // query 43 {44 int x, y; read (x); read (y); 45 ans = (ans * 2 + query (x, y) % mod; 46} 47 else // edge 48 {49 int x, y; read (x); read (y); 50 unionn (x, y ); 51} 52} 53 printf ("% d", ans); 54 return 0; 55}