目录
题意:给定 3 种操作,
第一种 1 u v 把 u 和 v 合并
第二种 2 l r 把 l - r 这一段区间合并
第三种 3 u v 判断 u 和 v 是不是在同一集合中。
析:很容易知道是用并查集来做,但是如果单纯的用并查集,肯定是要超时的,所以要用链表,如果合并了,就把链表指向,
这样就搞定了这个题。
代码如下:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 | #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair< int , int > P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 0x3f3f3f3f3f3f; const double PI = acos (-1.0); const double eps = 1e-8; const int maxn = 2e5 + 10; const int mod = 1000000007; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = { "0000" , "0001" , "0010" , "0011" , "0100" , "0101" , "0110" , "0111" , "1000" , "1001" , "1010" , "1011" , "1100" , "1101" , "1110" , "1111" }; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in( int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int p[maxn], nxt[maxn]; int Find( int x){ return x == p[x] ? x : p[x] = Find(p[x]); } int main(){ scanf ( "%d %d" , &n, &m); for ( int i = 0; i <= n; ++i) p[i] = i, nxt[i] = i + 1; while (m--){ int op, u, v; scanf ( "%d %d %d" , &op, &u, &v); if (op == 1){ int x = Find(u); int y = Find(v); if (x != y) p[y] = x; } else if (op == 2){ int x = Find(v); while ( true ){ if (u > v) break ; int y = Find(u); if (x != y) p[y] = x; int t = u; u = nxt[u]; nxt[t] = v+1; } } else printf ( "%s\n" , Find(u) == Find(v) ? "YES" : "NO" ); } return 0; } |
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步