Ysmmm的快乐小屋

A. 拜访约翰 找关系然后dijstra123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051#include <iostream>#include <queue> // BFS　　Ｉ．拜访约翰using namespace std;struct node { int x, y, v, c; bool operator<(const node& a) const { return v > a.v; }};priority_queue<node> q;int dx[4] = { 0, 1, 0, -1 }, dy[4] = { 1, 0, -1, 0 }, a[105][105], n, T;bool mk[105][105][3];int bfs() { while (!q.empty()) ...

A. 奶牛飞车 贪心去搞就好了12345678910111213141516171819const int N = 5e4 + 5;int a[N];int vis[N];int main(){ ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); int n , m , d , l; cin >> n >> m >> d >> l; for (int i = 1;i <= n ;i ++){ cin >> a[i]; } sort(a + 1 , a + n + 1); int tot = 1; int ans = 0; for (int i = 1; i <= n ; i ++){ if (a[i] - (ans / m * d) >= l) ++ ans; } cout << ans << endl; return 0;} B. 牛奶规划 1.反悔贪心123456 ...

A. 树 并查集 + dfs + 离线1234567891011121314151617181920212223242526272829303132333435363738394041const int N=1e5+2;char s[N][2];int x,y,a[N],b[N],fa[N],f[N];int n,m,cnt,hd[N],to[N],nxt[N],ans[N];void add(int x,int y){ to[++cnt]=y; nxt[cnt]=hd[x]; hd[x]=cnt;}void dfs(int x,int fat){ if(b[x])f[x]=x; else f[x]=f[fat]; fa[x]=fat; for(int i=hd[x];i;i=nxt[i]) dfs(to[i],x);}int find(int x){ return x==f[x]?x:f[x]=find(f[x]);}int main(){ scanf("%d%d",&n,&m ...

A. 闹市 在树上做差分， 求lca 最后再跑一遍dfs1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465const int N = 1e6+10;int n,k;struct{ int to,nex;}edge[N<<1];int head[N<<1],edge_num;inline void add(int x,int y){ edge[++edge_num].to = y; edge[edge_num].nex = head[x]; head[x] = edge_num;}int fa[N][25],dep[N];inline void dfs(int now,int fu){ fa[now][0] = fu; dep[now] = dep[fu]+1; for(int i=1;(1<<i)<=dep ...

A. 拆盒子 贪心贪过去就好了123456789101112131415signed main() { IOS; string s; cin >> s; int len = s.size(); int sum = 0; for (int i = 0; i < len; i++) { if (s[i] == '(') sum++; else if(s[i] == ')') sum--; else break; } cout << sum << endl; return 0;} B. 数组调整 枚举最小值，算最大值就好了123456789101112131415161718192021signed main() { IOS; int n; cin >> n; vector<int> v(n); for (int ...

A. 苹果树2 用线段树去维护dfs序的连续性1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374#include<bits/stdc++.h>#define M 200001using namespace std;int n,m,tot;int A[M],to[M];int L[M],R[M];vector<int>Q[M];class { public: int l,r,v;} tree[M<<2];void dfs(int x,int q) { L[x]=++tot,to[tot]=x; for(int i=0; i<Q[x].size(); ++i) { int y=Q[x][i]; if(y==q)continue; dfs(y,x); } R[x]=to ...

A. 俄罗斯国旗 直接模拟就好了12345678910111213141516171819202122232425262728293031#define MAX 55int W[MAX], B[MAX], R[MAX];string mp[MAX];signed main() { IOS; int n, m; cin >> n >> m; for (int i = 1; i <= n; i++) { cin >> mp[i]; mp[i] = ' ' + mp[i]; } for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { W[i] += mp[i][j] != 'W'; B[i] += mp[i][j] != 'B'; R[i] += mp[i][j] != 'R'; } W[i] += W[i - ...

123456789101112void tarjan(int u,int eid) { dfn[u]=low[u]=++indx; stk[++tp]=u; for(int i=head[u]; ~i; i=edge[i].nxt) { int v=edge[i].to; if(dfn[v]==0) { tarjan(v,i); low[u]=min(low[u],low[v]); } if(i!=(eid^1)) low[u]=min(low[u],dfn[v]); }} A. 桥（割边）- 1234567891011121314151617181920212223242526272829303132333435363738const int N=3e4+10;int n,m,a,b;vector<pair<int,int> >g[N];int dfn[N],low[N],cnt,ans;void DFS(int u,int bef){ dfn[u]=low[u]=++cn ...

A. 三人三数 比一下就好了1234567891011121314151617181920212223signed main() { IOS; int ans = 0; int n, a, b, c, d, e, f; cin >> n >> a >> b >> c >> d >> e >> f; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { for (int k = 1; k <= n; k++) {#define ch(x, y) min(abs(x - y), n - abs(x - y)) if (ch(a, i) <= 2 && ch(b, j) <= 2 && ch(c, k) <= 2) { ...

A. 搬草啦12345678910111213141516171819202122signed main() { IOS; int n; cin >> n; vector<int> v(n + 1); int sum = 0; for (int i = 1; i <= n; i++) { cin >> v[i]; sum += v[i]; } int tmp = sum / n; int res = 0; for (int i = 1; i <= n; i++) { if (v[i] < tmp) { res += tmp - v[i]; } } cout << res << endl; return 0;} B. 电泳技术 直接模拟写就好了123456789101112 ...