若推遲 \(u\) 必推遲 \(v\)，則連邊 <\(u,v\)>。
求強聯通分量後縮點，答案顯然是出度為 \(0\) 且 size 最小的 scc。

#include <iostream>
#include <cstdio>
using namespace std;
int n, m, h, a[100005], dfn[100005], uu, vv, loo[100005], bel[100005], scc, hav[100005], sta[100005], din, tot;
int hea[100005], cnt, ru[100005], ans=0x3f3f3f3f, chu[100005], minn;
bool
 
 ins[100005];
struct Edge{
    int too, nxt;
}edge[200005];
void add_edge(int fro, int too){
    edge[++cnt].nxt = hea[fro];
    edge[cnt].too = too;
    hea[fro] = cnt;
}
void tarjan(int x){
    sta[++din] = x;
    ins[x] = true;
    dfn[x] = loo[x] = ++tot;
    for(int i=hea[x]; i; i=edge[i].nxt){
        int
 
 t=edge[i].too;
        if(!dfn[t]){
            tarjan(t);
            loo[x] = min(loo[x], loo[t]);
        }
        else if(ins[t])
            loo[x] = min(loo[x], dfn[t]);
    }
    if(dfn[x]==loo[x]){
        int j;
        scc++;
        do{
            j = sta[din--];
            bel[j] = scc;
            hav[scc]++;
            ins[j] = false
 
;
        }while(dfn[j]!=loo[j]);
    }
}
int main(){
    cin>>n>>m>>h;
    for(int i=1; i<=n; i++)
        scanf("%d", &a[i]);
    for(int i=1; i<=m; i++){
        scanf("%d %d", &uu, &vv);
        if((a[uu]+1)%h==a[vv])  add_edge(uu, vv);
        if((a[vv]+1)%h==a[uu])  add_edge(vv, uu);
    }
    for(int i=1; i<=n; i++)
        if(!dfn[i])
            tarjan(i);
    for(int i=1; i<=n; i++)
        for(int j=hea[i]; j; j=edge[j].nxt){
            int t=edge[j].too;
            if(bel[i]!=bel[t])
                chu[bel[i]]++;
        }
    for(int i=1; i<=scc; i++)
        if(!chu[i] && ans>hav[i]){
            ans = hav[i];
            minn = i;
        }
    cout<<ans<<endl;
    for(int i=1; i<=n; i++)
        if(bel[i]==minn)
            printf("%d ", i);
    printf("\n");
    return 0;
}

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