HDU 4694 Important Sisters

题目大意：

给你N个点和M条边的有向图，其中第N个点是源点

让你求每个节点#I关于点#N的关键点的编号和

解题思路：

这题是2013 Multi-University Training Contest 9的题目，官方题解是用bitset或者floyd乱搞

然而丁神告诉我们这题是一题裸的Lengauer_Tarjan算法，具体算法见Tarjan论文，有详细的伪代码

附上丁神的模板

Lengauer-Tarjan algorithm的作用是求出dominator tree，这棵树上的每一个节点都是其儿子的idom

明显我们求出这棵树之后，原题要求的就是每个节点到根的路径上的各节点的编号和

O(MlogM)的算法和O(ma(N,M))的优化在于EVAL和LINK函数

succ数组存的是原图

fa数组存的是i结点的先驱，在dfs生成树上的父亲

dfn数组存的是i结点的新编号，redfn存的是i结点的原编号

prod数组存根据dfs重新排序之后的图

semi数组存半必经点的新编号，表示的是在dfs树上，节点i的祖先中，可以通过一系列的非树边走到i的，深度最小的祖先，i的直系父亲也可以是半必经点

bucket数组存的是，以i作为半必经点的点

idom数组存的是，i的immediate dominator

anc数组：step3里把点都先当成孤立的森林，然后每访问一个点，就将他和他父亲连边，anc数组存的就是结点的父           亲，并且compress用了类似并查集的方式压缩，以此更快速地找到深度最小的祖先

<pre name="code" class="cpp">#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <algorithm>#include <vector>#include <queue>#include <deque>#include <utility>#include <map>#include <set>#include <cctype>#include <climits>#include <stack>#include <cmath>#include <bitset>#include <numeric>#include <functional>using namespace std;vector<int> succ[50010], prod[50010], bucket[50010], dom_t[50010];int semi[50010], anc[50010], idom[50010], best[50010], fa[50010];int dfn[50010], redfn[50010];int timestamp;void dfs(int now){    dfn[now] = ++timestamp;    redfn[timestamp] = now;    anc[timestamp] = idom[timestamp] = 0;    semi[timestamp] = best[timestamp] = timestamp;    int sz = succ[now].size();    for(int i = 0; i < sz; ++i)    {        if(dfn[succ[now][i]] == -1)        {            dfs(succ[now][i]);            fa[dfn[succ[now][i]]] = dfn[now];        }        prod[dfn[succ[now][i]]].push_back(dfn[now]);    }}void compress(int now){    if(anc[anc[now]] != 0)    {        compress(anc[now]);        if(semi[best[now]] > semi[best[anc[now]]])            best[now] = best[anc[now]];        anc[now] = anc[anc[now]];    }}int eval(int now){    if(anc[now] == 0)        return now;    compress(now);    return best[now];}void debug(){    for(int i=timestamp;i>1;i--)        cout<<redfn[i]<<" "<<redfn[anc[i]]<<" "<<redfn[fa[i]]<<endl;    cout<<"---------------------"<<endl;}void lengauer_tarjan(int n){    memset(dfn, -1, sizeof dfn);    memset(fa, -1, sizeof fa);//memset(anc, 0, sizeof anc);//memset(idom, 0, sizeof idom);//for(int i = 0; i <= n; ++i)//best[i] = semi[i] = i;    timestamp = 0;    dfs(n);    fa[1] = 0;    for(int w = timestamp; w > 1; --w)    {        if(fa[w] == -1)            continue;        int sz = prod[w].size();        for(int i = 0; i < sz; ++i)        {            int u = eval(prod[w][i]);            if(semi[w] > semi[u])                semi[w] = semi[u];        }        debug();        bucket[semi[w]].push_back(w);        anc[w] = fa[w];        if(fa[w] == 0)            continue;        sz = bucket[fa[w]].size();        for(int i = 0; i < sz; ++i)        {            int u = eval(bucket[fa[w]][i]);            if(semi[u] < fa[w])                idom[bucket[fa[w]][i]] = u;            else                idom[bucket[fa[w]][i]] = fa[w];        }        bucket[fa[w]].clear();    }    for(int w = 2; w <= n; ++w)    {        if(fa[w] == -1)            continue;        if(idom[w] != semi[w])            idom[w] = idom[idom[w]];    }    idom[1] = 0;    for(int i = timestamp; i > 1; --i)    {        if(fa[i] == -1)            continue;        dom_t[idom[i]].push_back(i);    }}long long ans[50010];void get_ans(int now){    ans[redfn[now]] += redfn[now];    int sz = dom_t[now].size();    for(int i = 0; i < sz; ++i)    {        ans[redfn[dom_t[now][i]]] += ans[redfn[now]];        get_ans(dom_t[now][i]);    }}void MAIN(int n, int m){    for(int i = 0; i <= n; ++i)        succ[i].clear(), prod[i].clear(), bucket[i].clear(), dom_t[i].clear();    for(int i = 0, u, v; i < m; ++i)    {        scanf("%d%d", &u, &v);        succ[u].push_back(v);    }    lengauer_tarjan(n);    memset(ans, 0, sizeof ans);    get_ans(1);    for(int i = 1; i <= n; ++i)        printf("%I64d%c", ans[i], i == n ? '\n' :' ');}int main(){    int n, m;    while(scanf("%d%d", &n, &m) > 0)        MAIN(n, m);    return 0;}

O(ma（N,M）)模板来源丁神

</pre><pre name="code" class="cpp">//      whn6325689//Mr.Phoebe//http://blog.csdn.net/u013007900#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define eps 1e-9#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LLINF 1LL<<62template<class T>inline bool read(T &n){    T x = 0, tmp = 1;    char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------//Auther:winoros ding//input : succ//output dom_t and idom redfn//notice that the index i in dom_t[i] and dom_t[i][j] is the vertex's timestamp in dfs//hence you need redfn[i] to find the original vertex//o(mlogm) where m is the number of edges//UPD:new version is o(m|á(m, n)), the previous version is in commentconst int vector_num=50000; //max number of verticesvector<int> succ[vector_num+10], prod[vector_num+10], bucket[vector_num+10], dom_t[vector_num+10];int semi[vector_num+10], anc[vector_num+10], idom[vector_num+10], best[vector_num+10], fa[vector_num+10];int dfn[vector_num+10], redfn[vector_num+10];int child[vector_num+10], size[vector_num+10];int timestamp;void dfs(int now){    dfn[now] = ++timestamp;    redfn[timestamp] = now;    anc[timestamp] = idom[timestamp] = child[timestamp] = size[timestamp] = 0;    semi[timestamp] = best[timestamp] = timestamp;    int sz = succ[now].size();    for(int i = 0; i < sz; ++i)    {        if(dfn[succ[now][i]] == -1)        {            dfs(succ[now][i]);            fa[dfn[succ[now][i]]] = dfn[now];        }        prod[dfn[succ[now][i]]].push_back(dfn[now]);    }}void compress(int now){    if(anc[anc[now]] != 0)    {        compress(anc[now]);        if(semi[best[now]] > semi[best[anc[now]]])            best[now] = best[anc[now]];        anc[now] = anc[anc[now]];    }}inline int eval(int now){    if(anc[now] == 0)        return now;    else    {        compress(now);        return semi[best[anc[now]]] >= semi[best[now]] ? best[now]               : best[anc[now]];    }}inline void link(int v, int w){    int s = w;    while(semi[best[w]] < semi[best[child[w]]])    {        if(size[s] + size[child[child[s]]] >= 2*size[child[s]])        {            anc[child[s]] = s;            child[s] = child[child[s]];        }        else        {            size[child[s]] = size[s];            s = anc[s] = child[s];        }    }    best[s] = best[w];    size[v] += size[w];    if(size[v] < 2*size[w])        swap(s, child[v]);    while(s != 0)    {        anc[s] = v;        s = child[s];    }}void lengauer_tarjan(int n)   // n is the vertices' number{    memset(dfn, -1, sizeof dfn);    memset(fa, -1, sizeof fa);    timestamp = 0;    dfs(n);    fa[1] = 0;    for(int w = timestamp; w > 1; --w)    {        int sz = prod[w].size();        for(int i = 0; i < sz; ++i)        {            int u = eval(prod[w][i]);            if(semi[w] > semi[u])                semi[w] = semi[u];        }        bucket[semi[w]].push_back(w);        //anc[w] = fa[w]; link operation for o(mlogm) version        link(fa[w], w);        if(fa[w] == 0)            continue;        sz = bucket[fa[w]].size();        for(int i = 0; i < sz; ++i)        {            int u = eval(bucket[fa[w]][i]);            if(semi[u] < fa[w])                idom[bucket[fa[w]][i]] = u;            else                idom[bucket[fa[w]][i]] = fa[w];        }        bucket[fa[w]].clear();    }    for(int w = 2; w <= timestamp; ++w)    {        if(idom[w] != semi[w])            idom[w] = idom[idom[w]];    }    idom[1] = 0;    for(int i = timestamp; i > 1; --i)    {        if(fa[i] == -1)            continue;        dom_t[idom[i]].push_back(i);    }}long long ans[50010];void get_ans(int now){    ans[redfn[now]] += redfn[now];    int sz = dom_t[now].size();    for(int i = 0; i < sz; ++i)    {        ans[redfn[dom_t[now][i]]] += ans[redfn[now]];        get_ans(dom_t[now][i]);    }}void init(int n, int m){    for(int i=0; i<=n; i++)        succ[i].clear(), prod[i].clear(), bucket[i].clear(), dom_t[i].clear();    CLR(ans,0);}int main(){    int n, m;    while(read(n)&&read(m))    {        init(n,m);        for(int i=0,u,v; i<m; i++)        {            read(u), read(v);            succ[u].push_back(v);        }        lengauer_tarjan(n);        get_ans(1);        for(int i=1; i<=n; i++)            printf("%I64d%c", ans[i], i == n ? '\n' :' ');    }    return 0;}