codefores741A Arpa's loud Owf and Mehrdad's evil plan(图找环)

题意：

对于给定的一个数列a[n],如果现在点i，那么下一个点就在a[i]，现在问给定n个点，找到一个最小的t,使得对于从任意一个点x出发，经过t次之后，到达的点y, 使得点y经过t次之后会回到点x。

要点：

一看就是图找环，找到所有独立的环对应的节点数，对所有节点数求最小公倍数就是结果。有个地方要注意就是，当节点数是偶数时，比如1->2->3->4->1，这种情况要将节点数/2计算，因为比如1到达3要2次，3到达1也是两次，肯定比从1绕一圈再到1要小，奇数没有办法。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 105;
int a[N], vis[N];
ll b[N];int dfs(int u, int s, int l)//有向图找环
{if (u == s) {l = l + 1;return l;}if (vis[u] && u != s) {return -1;}vis[u] = 1;if (a[u] == s)return dfs(a[u], s, l);elsereturn dfs(a[u], s, l + 1);
}
ll gcd(ll a, ll b)
{if (b == 0) return a;return gcd(b, a%b);
}int main()
{int n;scanf("%d", &n);for (int i = 1; i <= n; i++)scanf("%d", &a[i]);memset(vis, 0, sizeof(vis));int cnt = 0;bool flag = true;for (int i = 1; i <= n; i++){if (vis[i])continue;int temp = dfs(a[i], i, 1);if (temp != -1){if (temp % 2 == 0)temp /= 2;b[cnt++] = temp;}else{flag = false;break;}}if (!flag)printf("-1\n");else{ll lcm, c = b[0], e;for (int i = 0; i < cnt; i++)//计算lcm{e = gcd(c, b[i]);lcm = c*b[i] / e;c = lcm;}printf("%I64d\n", lcm);}return 0;
}

转载于:https://www.cnblogs.com/seasonal/p/10343651.html

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