【LCA】 POJ 1470 Closest Common Ancestors
给出一棵树 再给出m个询问 求m个询问后
最近每个点成为公共祖先的次数 (输出大于0次的)
/* * LCA离线算法,Tarjan * 复杂度O(n+Q); */ #include <cstdio> #include <cstring> #include <cstdlib> #include <string> #include <iostream> #include <algorithm> #include <sstream> #include <cmath> using namespace std; #include <queue> #include <stack> #include <vector> #include <deque> #define cler(arr, val) memset(arr, val, sizeof(arr)) #define FOR(i,a,b) for(int i=a;i<=b;i++) #define IN freopen ("in.txt" , "r" , stdin); #define OUT freopen ("out.txt" , "w" , stdout); typedef long long LL; const int MAXN = 1000+5; const int MAXM = 550000; const int INF = 0x3f3f3f3f; const int mod = 1000000007; int F[MAXN];//需要初始化为-1 int find(int x) { if(F[x] == -1)return x; return F[x] = find(F[x]); } void bing(int u,int v) { int t1 = find(u); int t2 = find(v); if(t1 != t2) F[t1] = t2; } //************************ bool vis[MAXN];//访问标记 int ancestor[MAXN];//祖先 struct Edge { int to,next; } edge[MAXN*2]; int head[MAXN],tot; void addedge (int u,int v) { edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } struct Query { int q,next; int index;//查询编号 } query[MAXM*2]; int answer[MAXM];//存储最后的查询结果,下标0~Q-1 int h[MAXM]; int tt; int Q; void add_query(int u,int v,int index) { query[tt].q = v; query[tt].next = h[u]; query[tt].index = index; h[u] = tt++; query[tt].q = u; query[tt].next = h[v]; query[tt].index = index; h[v] = tt++; } void init() { tot = 0; memset(head,-1,sizeof (head)); tt = 0; memset(h,-1,sizeof(h)); memset(vis,false,sizeof(vis)); memset(F,-1,sizeof(F)); memset(ancestor,0,sizeof(ancestor)); } void LCA(int u) { ancestor[u] = u; vis[u] = true; for(int i = head[u]; i != -1; i = edge[i].next) { int v = edge[i].to; if(vis[v])continue; LCA(v); bing(u,v); ancestor[find(u)] = u; } for(int i = h[u]; i != -1; i = query[i].next) { int v = query[i]. q; if(vis[v]) { answer[query[i].index] = ancestor[find(v)]; } } } bool flag[MAXN]; int Count_num[MAXN]; int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); #endif int k,u,v; int n,m; while(scanf("%d",&n)!=EOF) { init(); cler(flag,false); for(int i=0; i<n; i++) { scanf("%d:(%d)",&u,&k); for(int i=0; i<k; i++) { scanf("%d",&v); flag[v]=true; addedge(u,v); addedge(v,u); } } scanf("%d",&m); for(int i=0; i<m; i++) { char s; cin>>s; scanf("%d %d)",&u,&v); add_query(u,v,i); } int root; for(int i=1; i<=n; i++) { if(!flag[i]) { root=i; break; } } LCA(root); cler(Count_num,0); for(int i=0; i<m; i++) Count_num[answer[i]]++; for(int i=1; i<=n; i++) if(Count_num[i]>0) printf("%d:%d\n",i,Count_num[i]); } return 0; }