若推迟 (u) 必推迟 (v)，则连边 <(u,v)>。
求强联通分量后缩点，答案显然是出度为 (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("
");
return 0;
}