poj 1470 Closest Common Ancestors （离线LCA Tarjan）

题意：求每个点作为lca的次数。

做法：裸的lca题，离线做法就是利用并查集，关键思想在于对于当前点的询问对应的另一个点已经被访问过了，那么那个点的祖先就是这个询问的lca。

AC代码：

//#pragma comment(linker, "/STACK:102400000,102400000")#include<cstdio>#include<ctype.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<cstdlib>#include<stack>#include<queue>#include<set>#include<map>#include<cmath>#include<ctime>#include<string.h>#include<string>#include<sstream>#include<bitset>using namespace std;#define ll long long#define ull unsigned long long#define eps 1e-8#define NMAX 1000000000#define MOD 51123987#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define PI acos(-1)#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x.end())template<class T>inline void scan_d(T &ret){    char c;    int flag = 0;    ret=0;    while(((c=getchar())<'0'||c>'9')&&c!='-');    if(c == '-')    {        flag = 1;        c = getchar();    }    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();    if(flag) ret = -ret;}template<class T> inline T Max(T a, T b){ return a > b ? a : b; }template<class T> inline T Min(T a, T b){ return a < b ? a : b; }const int maxn = 1000+10;struct Edge{    int v,next;}e[maxn*2];int head[maxn],nct;inline void add_edge(int u, int v){    e[nct].v = v; e[nct].next = head[u];    head[u] = nct++;}vector<int>v[maxn];bool vis[maxn],cnt[maxn];int ans[maxn],fa[maxn];int findit(int x){    int k = x;    while(fa[k] != k)        k = fa[k];    while(fa[x] != k)    {        int tmp = fa[x];        fa[x] = k;        x = tmp;    }    return k;}void dfs(int u){    for(int i = head[u]; i != -1; i = e[i].next)    {        int v = e[i].v;        dfs(v);        fa[v] = u;    }    vis[u] = 1;    int sz = v[u].size();    for(int i = 0; i < sz; i++)        if(vis[v[u][i]])            ans[findit(v[u][i])]++;}int main(){#ifdef GLQ    freopen("input.txt","r",stdin);//    freopen("o.txt","w",stdout);#endif    int n;    while(~scanf("%d",&n))    {        memset(head,-1,sizeof(head));        nct = 0;        for(int i = 1; i <= n; i++)        {            cnt[i] = vis[i] = false;            ans[i] = 0;            fa[i] = i;            v[i].clear();        }        for(int i = 1; i <= n; i++)        {            int u,v,t1;            char c;            scan_d(u);            scan_d(t1);            if(t1 == 0) continue;            for(int j = 1; j <= t1; j++)            {                scan_d(v);                add_edge(u,v);                cnt[v] = true;            }        }        int t1;        scan_d(t1);        for(int i = 1; i <= t1; i++)        {            int x,y;            char c;            while((c = getchar()) != '(');            scanf("%d%d",&x,&y);            v[x].push_back(y);            v[y].push_back(x);        }        while(getchar() != ')');        for(int i = 1; i <= n; i++)            if(cnt[i] == false)            {                dfs(i);                break;            }        for(int i = 1; i <= n; i++)            if(ans[i] != 0) printf("%d:%d\n",i,ans[i]);    }    return 0;}