Day-0

Day-0

T1

• 从一点开始以固定方向行走，会出现循环节

  • 该结论由 $n=m$ 的测试点推出

• 判断两个串是否相同

  • 字符串哈希

• 倍增预处理出长度为 $2^{len}$​ 的循环节子串

T2

• 最小环计数

• Floyd 会超时

• 计数由 $x,y,st$ 构成的环

  • 依题解代码

• 同学赛时代码

  • 更好理解

  #include <bits/stdc++.h>
  #define N 6005  //记得改数据范围！！！！！！！！！！！！！！！！！！ !!!!!!!!!!
  #define M 10000005
  #define ll long long
  #define lowbit(x) (x & -x)
  #define ls(x) llll[x]
  #define rs(x) rrrr[x]
  #define x first
  #define y second
  #define PII pair<int, int>
  #define PIL pair<int, ll>
  #define PLI pair<ll, int>
  #define PLL pair<ll, ll>
  #define mk make_pair
  #define mod 998244353  //改模数！！！！！！！！！！！！
  #define bas 29
  //#define int long long
  using namespace std;
  int n, m;
   
  struct node {
      int v, id;
  };
  vector<node> mp[N];
  struct eee {
      int u, v;
  } edge[N];
  int dis[N], vis[N];
  ll cnt[N];
  signed main() {
      freopen("b.in", "r", stdin);
      freopen("b.out", "w", stdout);
      ios::sync_with_stdio(false);
      cin.tie(0), cout.tie(0);
      cin >> n >> m;
      for (int i = 1; i <= m; ++i) {
          int u, v;
          cin >> u >> v;
          edge[i] = (eee){ u, v };
          mp[u].push_back((node){ v, i });
          mp[v].push_back((node){ u, i });
      }
      int minn = 1e9;
      ll ans = 0;
      for (int i = 1; i <= m; ++i) {
          int s = edge[i].u, t = edge[i].v;
          queue<PII> l;
          l.push(mk(s, 0));
          memset(vis, 0, sizeof(vis));
          memset(dis, 0x3f, sizeof(dis));
          memset(cnt, 0, sizeof(cnt));
          cnt[s] = 1;
          dis[s] = 0;
          while (!l.empty()) {
              int u = l.front().x;
              l.pop();
              if (vis[u])
                  continue;
              vis[u] = 1;
              for (int j = 0; j < mp[u].size(); ++j) {
                  int v = mp[u][j].v;
                  if (mp[u][j].id > i) {
                      l.push(mk(v, dis[u] + 1));
                      if (dis[u] + 1 < dis[v]) {
                          dis[v] = dis[u] + 1;
                          cnt[v] = cnt[u];
                      } else if (dis[u] + 1 == dis[v])
                          cnt[v] += cnt[u];
                  }
              }
          }
          if (dis[t] + 1 < minn) {
              minn = dis[t] + 1;
              ans = cnt[t];
          } else if (dis[t] + 1 == minn)
              ans += cnt[t];
      }
      cout << ans;
      return 0;
  }

T3

• 区间DP

  • 适用范围：子问题解决 能使得 母问题解决

• std::max

  • 传入多个参数

  • 好像更快

    # include <bits/stdc++.h>
    # define int long long
    using namespace std;
     
    signed main(){
    	int a = 6, b = 66, c = 666;
    	int maxi = max({a, b, c});
    	cout << maxi << "\n";
    }

  //6.838S
  # include <bits/stdc++.h>
  # define int long long
  using namespace std;
  const int N = (int)5e7 + 10;
   
  int n = (int)5e7;
  int a[N];
   
  signed main(){
  	freopen("1.in", "r", stdin);
  	ios::sync_with_stdio(0);
  	cin.tie(0); cout.tie(0);
  	for(int i = 1; i <= n; i++){
  		cin >> a[i];
  	}
  	int maxi = 0;
  	for(int i = 1; i <= n; i += 6){
  		maxi = max(a[i], max(a[i + 1], max(a[i + 2], max(a[i + 3], max(a[i + 4], a[i + 5])))));
  	}
  	cout << maxi << "\n";
  }

  //6.711S
  # include <bits/stdc++.h>
  # define int long long
  using namespace std;
  const int N = (int)5e7 + 10;
   
  int n = (int)5e7;
  int a[N];
   
  signed main(){
  	freopen("1.in", "r", stdin);
  	ios::sync_with_stdio(0);
  	cin.tie(0); cout.tie(0);
  	for(int i = 1; i <= n; i++){
  		cin >> a[i];
  	}
  	int maxi = 0;
  	for(int i = 1; i <= n; i += 6){
  		maxi = max({a[i], a[i + 1], a[i + 2], a[i + 3], a[i + 4], a[i + 5]});
  	}
  	cout << maxi << "\n";
  }

T4

• 重点在于将原题建图
  • 转化为同一连通块内的问题
  • 将城市看作点，奇数个入度的点需要带仆人
  • 同一连通块内点的入度可以均摊