【TP SRM 703 div2 500】 GCDGraph

Problem Statement

You are given four ints: n, k, x, and y. The ints n and k describe a simple undirected graph. The graph has n nodes, numbered 1 through n. Two distinct vertices i and j are connected by an edge if and only if gcd(i, j) > k. Here, gcd(i, j) denotes the greatest common divisor of i and j. The ints x and y are the numbers of two (not necessarily distinct) vertices in our graph. Return “Possible” if it is possible to travel from x to y by following the edges of our graph. Otherwise, return “Impossible”. In other words, return “Possible” if our graph contains a path that connects the nodes x and y, and “Impossible” if there is no such path.

Definition

Class:

GCDGraph

Method:

possible

Parameters:

int, int, int, int

Returns:

string

Method signature:

string possible(int n, int k, int x, int y)

(be sure your method is public)

Limits

Time limit (s):

2.000

Memory limit (MB):

256

Stack limit (MB):

256

Constraints

n will be between 2 and 1,000,000, inclusive.

k will be between 0 and n, inclusive.

x and y will be between 1 and n, inclusive.

Examples

0)

12

2

8

9

Returns: “Possible”

We have a graph with n = 12 nodes. As k = 2, vertices i and j are connected by an edge if and only if gcd(i, j) is strictly greater than 2. In this graph it is possible to travel from node 8 to node 9. One possible path: 8 -> 4 -> 12 -> 9.

1)

12

2

11

12

Returns: “Impossible”

This is the same graph as in Example 0, but now we are interested in another pair of nodes. It is not possible to travel from node 11 to node 12. In particular, in our graph node 11 is an isolated node because for any other node x we have gcd(11, x) = 1.

2)

12

2

11

11

Returns: “Possible”

A node is always reachable from itself.

3)

10

2

8

9

Returns: “Impossible”

4)

1000000

1000

12345

54321

Returns: “Possible”

5)

1000000

2000

12345

54321

Returns: “Impossible”

6)

2

0

1

2

Returns: “Possible”

【题目链接】:

【题解】



大意是说两个节点之间有边当且仅当两个节点的标号的gcd>k;

可以这样.

先枚举比k大的且比n小的数i;

然后它的倍数和它之间连了一条边.

表示这两个数的最大公因数为i;而i大于k;所以满足题意;

而所有i的出度点之间则肯定也有路径可以到达了。

可以这样想？

两个数x,y的gcd为i

则i和y的gcd为i

i和x的gcd也为i

即x和y肯定是i的倍数.

所以如果i大于k

这对关系x,y肯定能找出来;(用并查集判断就可以了);

其他的要通过间接关系找出来的也同理吧！

用并查集描述两个数之间是否联通即可.



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 1e6+100;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);

int f[MAXN];

int ff(int x)
{
    if (f[x]!=x)
        f[x] = ff(f[x]);
    else
        return x;
    return f[x];
}

class GCDGraph
{
    public:
        string possible(int n, int k, int x, int y)
        {
            rep1(i,1,n)
                f[i] = i;
            rep1(i,k+1,n)
            {
                int fa = ff(i);
                for (int j = 2*i;j <= n;j+=i)
                {
                    int r1 = ff(j);
                    f[r1] = fa;
                }
            }
            if (ff(x)==ff(y))
                return "Possible";
            else
                return "Impossible";
        }
};