[NOI2008]假面舞会——数论+dfs找环

原题戳这里

思路

分三种情况讨论：

1.有环

那显然是对于环长取个\(gcd\)

2.有类环

也就是这种情况

1→2→3→4→5→6→7，1→8→9→7

假设第一条链的长度为\(l_1\)，第二条为\(l_2\)，那么\(l_1\)和\(l_2\)需要满足\(l_1\equiv l_2(mod\ k)\)，也就是\(k|(l_1-l_2)\)。如果我们建权值为\(-1\)的反向边的话，找出来的环就涵盖了这种情况，并且取\(gcd\)就能满足等式

3.有链

对答案无影响

最后还需要加一个特判，就是只有链的情况

具体可以看代码

#include <algorithm>
#include  <iostream>
#include   <cstdlib>
#include   <cstring>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <cmath>
#include     <ctime>
#include     <queue>
#include       <map>
#include       <set>

using namespace std;

#define IINF 0x3f3f3f3f3f3f3f3fLL
#define ull unsigned long long
#define pii pair<int, int>
#define uint unsigned int
#define mii map<int, int>
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back

#define N 100000
#define M 1000000

struct Edge {
  int next, to, w;
}e[2*M+5];

int n, m, maxans, minans;
int fa[N+5], head[N+5], eid, vis[N+5], d[N+5];
int mmax[N+5], mmin[N+5];

int find(int x) {
  return fa[x] == x ? x : fa[x] = find(fa[x]);
}

int gcd(int a, int b) {
  return !b ? a : gcd(b, a%b);
}

void addEdge(int from, int to, int w) {
  e[++eid].next = head[from];
  e[eid].to = to;
  e[eid].w = w;
  head[from] = eid;
}

void dfs(int u, int x) {
  if(vis[u]) {
    maxans = gcd(abs(x-d[u]), maxans);
    return ;
  }
  vis[u] = 1;
  mmin[find(u)] = min(mmin[find(u)], x);
  mmax[find(u)] = max(mmax[find(u)], x);
  d[u] = x;
  for(int i = head[u]; i; i = e[i].next) {
    int v = e[i].to, w = e[i].w;
    dfs(v, x+w);
  }
}

int main() {
  scanf("%d%d", &n, &m);
  for(int i = 1; i <= n; ++i) fa[i] = i;
  for(int i = 1, x, y; i <= m; ++i) {
    scanf("%d%d", &x, &y);
    addEdge(x, y, 1), addEdge(y, x, -1);
    int fx = find(x), fy = find(y);
    if(fx != fy) fa[fy] = fx;
  }
  memset(mmin, 0x3f, sizeof mmin);
  for(int i = 1; i <= n; ++i) {
    if(vis[i]) continue;
    dfs(i, 0);
  }
  for(int i = 3; i <= maxans; ++i) {
    if(maxans%i == 0) {
      minans = i;
      break;
    }
  }
  if(!maxans)
    for(int i = 1; i <= n; ++i)
      if(find(i) == i)
        maxans += mmax[i]-mmin[i]+1;
  if(!minans) minans = 3;
  if(maxans < 3) printf("-1 -1\n");
  else printf("%d %d\n", maxans, minans);
  return 0;
}