HDU4694：Important Sisters（支配树）

Description
There are N clones of Misaka Mikoto (sisters) forming the Misaka network. Some pairs of sisters are connected so that one of them can pass message to the other one. The sister with serial number N is the source of all messages. All the other sisters get message directly or indirectly from her. There might be more than one path from sister #N to sister #I, but some sisters do appear in all of these paths. These sisters are called important sister of sister #K. What are the important sisters of each sister?

Input
There are multiple test cases. Process to the End of File.
The first line of each test case contains two integers: the number of sisters 1 ≤ N ≤ 50,000 and the number of connections 0 ≤ M ≤ 100,000. The following M lines are M connections 1 ≤ Ai, Bi ≤ N, indicating that Ai can pass message to Bi.

Output
For each test case, output the sum of the serial numbers of important sisters of each sister, separated with single space.

Sample Input
3 2
3 2
2 1
5 7
3 2
1 2
2 1
3 1
3 2
5 3
5 4

Sample Output
6 5 3
9 10 8 9 5

题意：
给一张n个点m条边的有向图，求n号节点到每个点的必经点的编号之和。

题解：
支配树。
求出每个点的支配点之后，从dfn编号小的递推至大的

$$ans[i]=\left\{ \begin{aligned} & i \,\, (\,\,idom[i]=i\,\,)\\ & ans[\,idom[i]\,]+i\,\,(\,\,idom[i]!=i\,\,) \end{aligned} \right.$$

代码：

#include<bits/stdc++.h>
using namespace std;
inline int rd(){
    char ch=getchar();int i=0,f=1;
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){i=(i<<1)+(i<<3)+ch-'0';ch=getchar();}
    return i*f;
}
inline void W(long long x){
    static int buf[50];
    if(!x){putchar('0');return;}
    if(x<0){putchar('-');x=-x;}
    while(x){buf[++buf[0]]=x%10;x/=10;}
    while(buf[0])putchar(buf[buf[0]--]+'0');
}
const int N=1e5+50,M=4e5+50;
int n,m;
struct Graph{
    int dfn[N],g[N],fa[N],id[N],v[M],nt[M],ec,tot;
    int sdom[N],idom[N],mnsd[N],anc[N];long long sum[N];
    vector<int>pre[N],buc[N];
    inline void getanc(int p){
        if(anc[p]!=p)getanc(anc[p]);
        if(dfn[sdom[mnsd[anc[p]]]]<dfn[sdom[mnsd[p]]])mnsd[p]=mnsd[anc[p]];
        anc[p]=anc[anc[p]];
    }
    inline int eval(int p){
        getanc(p);
        return mnsd[p];
    }
    inline void clear(){
        memset(dfn,0,sizeof(dfn));
        memset(g,0,sizeof(g));
        memset(anc,0,sizeof(anc));
        memset(sum,0,sizeof(sum));
        ec=0;tot=0;
    }
    inline void add(int x,int y){nt[++ec]=g[x];g[x]=ec;v[ec]=y;}
    inline void dfs(int now,int f=0){
        fa[now]=f;dfn[now]=++tot;id[tot]=now;
        sdom[now]=idom[now]=mnsd[now]=anc[now]=now;
        for(int j=g[now];j;j=nt[j]){
            if(dfn[v[j]])continue;
            dfs(v[j],now);
        }
    }
    inline void solve(){
        for(int i=1;i<=n;i++)pre[i].clear(),buc[i].clear();
        for(int i=1;i<=m;i++){
            int x=rd(),y=rd();
            add(x,y);
            pre[y].push_back(x);
        }
        dfs(n,0);
        for(int i=tot;i>=2;i--){
            int u=id[i];
            for(int j=pre[u].size()-1;j>=0;j--){
                int t=eval(pre[u][j]);if(!dfn[t])continue;
                if(dfn[sdom[u]]>dfn[sdom[t]])sdom[u]=sdom[t];
            }
            int f=fa[u];
            anc[u]=f;
            buc[sdom[u]].push_back(u);
            for(int j=buc[f].size()-1;j>=0;j--){
                int v=buc[f][j],t=eval(v);
                if(dfn[sdom[t]]<dfn[sdom[v]])idom[v]=t;
                else idom[v]=sdom[v];
            }
            buc[f].clear();
        }
        for(int i=2;i<=tot;i++)
            if(idom[id[i]]!=sdom[id[i]])idom[id[i]]=idom[idom[id[i]]];
        sum[id[1]]=id[1];
        for(int i=2;i<=tot;i++)
            sum[id[i]]=sum[idom[id[i]]]+id[i];
        for(int i=1;i<n;i++)W((dfn[i]?sum[i]:0)),putchar(' ');
        W((dfn[n]?sum[n]:0));
        putchar('\n');
    }
}g;
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        g.solve();
        g.clear();
    }
}