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分析
因为区间会被后面的区间覆盖,所以考虑从后往前处理
我们用 f [ i ] f[i] f[i]表示 i i i节点前第一个没有被染色的节点,然后用并查集进行维护即可
代码
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e6 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int f[N];
int n,m,p,q;
int x[N],y[N],z[N];
int ans[N];int find(int x){if(x != f[x]) f[x] = find(f[x]);return f[x];
}int main() {read(n),read(m),read(p),read(q);for(int i = 1;i <= n;i++) f[i] = i;for(int i = m;i;i--){int l = (i * p + q) % n + 1,r = (i * q + p) % n + 1;if(l > r) swap(l,r);for(int j = r;j >= l;j = f[j]){int x = find(j);if(x == j) ans[j] = i,f[j] = find(j - 1);}}for(int i = 1;i <= n;i++) di(ans[i]);return 0;
}
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