[HDU4694]Important Sisters

Description

在御坂网络里有N<=5*10^4个炮姐，之间存在单向的信息传递关系，构成一张M<=10^5的有向图
炮姐N是所有消息的源头
如果去掉炮姐b，炮姐N就无法给炮姐a传递信息，则称b是a的important sister
求每个炮姐的所有important sister的编号之和

Solution

支配树裸题，每个点的答案即为它在支配树上到根的点的编号之和。
注意图不连通的情况要输出0

Code

#include <vector>#include <cstdio>#include <cstring>#include <algorithm>#define fo(i,a,b) for(int i=a;i<=b;i++)#define fd(i,a,b) for(int i=a;i>=b;i--)#define rep(i,a) for(int i=lst[a];i;i=nxt[i])using namespace std;typedef long long ll;int read() {    char ch;    for(ch=getchar();ch<'0'||ch>'9';ch=getchar());    int x=ch-'0';    for(ch=getchar();ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';    return x;}void write(ll x) {    if (!x) {putchar('0');return;}    char ch[20];int tot=0;    for(;x;x/=10) ch[++tot]=x%10+'0';    fd(i,tot,1) putchar(ch[i]);}const int N=5*1e4+5;typedef vector<int> vec;#define pb(a) push_back(a)vec pre[N],dom[N];int lst[N],nxt[N<<1],t[N<<1],l;void add(int x,int y) {    t[++l]=y;nxt[l]=lst[x];lst[x]=l;}int n,m,x,y,semi[N],idom[N];int id[N],dfn[N],fa[N],tot;void dfs(int x) {    dfn[x]=++tot;id[tot]=x;    rep(i,x) {        pre[t[i]].pb(x);        if (!dfn[t[i]]) {            fa[t[i]]=x;            dfs(t[i]);        }     }}int father[N],val[N];int get(int x) {    if (father[x]==x) return x;    int y=get(father[x]);    if (dfn[semi[val[father[x]]]]<dfn[semi[val[x]]]) val[x]=val[father[x]];    return father[x]=y;}int smin(int x,int y) {return dfn[x]<dfn[y]?x:y;}void solve() {    fd(i,tot,2) {        int x=id[i];        if (!pre[x].empty())            fo(j,0,pre[x].size()-1)                if (dfn[pre[x][j]]<dfn[x]) semi[x]=smin(semi[x],pre[x][j]);                else {                    get(pre[x][j]);                    semi[x]=smin(semi[x],semi[val[pre[x][j]]]);                }        father[x]=fa[x];dom[semi[x]].pb(x);        if (!dom[fa[x]].empty())            fo(j,0,dom[fa[x]].size()-1) {                int v=dom[fa[x]][j];get(v);                int u=val[v];                idom[v]=(dfn[semi[u]]<dfn[semi[v]])?u:fa[x];            }    }    fo(i,2,tot) {        int x=id[i];        if (idom[x]!=semi[x]) idom[x]=idom[idom[x]];    }}ll ans[N];void calc() {    fo(i,1,n) ans[i]=0;    fo(i,1,tot) {        int x=id[i];        ans[x]+=x;        if (idom[x]) ans[x]+=ans[idom[x]];    }}int main() {    //freopen("dominator-tree.in","r",stdin);    //freopen("dominator-tree.out","w",stdout);    for(;scanf("%d%d",&n,&m)!=EOF;) {        fo(i,1,n) lst[i]=0;l=0;        fo(i,1,n) pre[i].clear();        fo(i,1,n) dom[i].clear();        fo(i,1,m) {            x=read();y=read();            add(x,y);        }        fo(i,1,n) dfn[i]=idom[i]=0;tot=0;        fo(i,1,n) father[i]=val[i]=semi[i]=i;        dfs(n);solve();        calc();        fo(i,1,n) {            write(ans[i]);            if (i!=n) putchar(' ');        }        puts("");    }}