CodeForces 566D Restructuring Company (并查集+链表)

Posted 2020-09-28 dwtfukgv

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：

#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;
}