algzjh的博客

Nearest Common AncestorsTime Limit: 1000MS Memory Limit: 10000K Total Submissions: 30072 Accepted: 15378 DescriptionA rooted tree is a well-known data structure in computer science and en

int dep[N<<1];//记录欧拉序列每个位置的深度int fin[N];//树中每个点第一次访问时的idxint p[N<<1];//欧拉序列int dp[N<<1][20];//ST表int cnt;//idx//p[]、dep[]以及dp的第一维记得开点数两倍。//因为这三个数组的操作对象都是欧拉序列，而欧拉序列长度为2*n-1int min(int i,int j){

const int LOG = 20;//根据点数来取，取LOG >=logn就行了 20已经可以处理1e6以下的数据了int par[N][LOG],dep[N];void dfs(int u,int fa,int depth){ dep[u]=depth; if(u==1){ //对于根来说，往上跳无论几下都是他自己。我习惯把1定为根所以写 u == 1 for

Popular CowsTime Limit: 2000MS Memory Limit: 65536K Total Submissions: 35075 Accepted: 14295 DescriptionEvery cow’s dream is to become the most popular cow in the herd. In a herd of N (1

FFF at ValentineTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 417 Accepted Submission(s): 208Problem DescriptionAt Valentine’s eve, Shyloc

//树形dp求树的直径void dfs(int x){ vis[x]=1; for(int i=head[i];i!=-1;i=e[i].nxt){ int v = e[i].v; int w = e[i].w; if(vis[v]) continue; dfs(v); ans = max(ans,d...

int block,top,id;int low[MAXN],dfn[MAXN],ins[MAXN];int stk[MAXN],belong[MAXN],sz[MAXN];void init(){ top=0;block=0;id=0; memset(dfn,0,sizeof(dfn)); memset(ins,0,sizeof(ins));}void ta...