hdu2767 Proving Equivalences

先求一发 scc 然后缩点，然后根据入度出度为 (0) 的点的个数乱搞。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int T, n, m, hea[20005], cnt, uu, vv, bel[20005], scc, din, sta[20005], idx;
int dfn[20005], loo[20005], ind[20005], oud[20005], cnt1, cnt2;
bool ins[20005];
struct Edge{
	int too, nxt;
}edge[100005];
void add_edge(int fro, int too){
	edge[++cnt].nxt = hea[fro];
	edge[cnt].too = too;
	hea[fro] = cnt;
}
void tarjan(int x){
	dfn[x] = loo[x] = ++idx;
	ins[x] = true;
	sta[++din] = x;
	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]){
		scc++;
		int j;
		do{
			j = sta[din--];
			ins[j] = false;
			bel[j] = scc;
		}while(dfn[j]!=loo[j]);
	}
}
int main(){
	cin>>T;
	while(T--){
		memset(hea, 0, sizeof(hea));
		memset(dfn, 0, sizeof(dfn));
		memset(loo, 0, sizeof(loo));
		memset(ind, 0, sizeof(ind));
		memset(oud, 0, sizeof(oud));
		cnt = scc = din = idx = cnt1 = cnt2 = 0;
		scanf("%d %d", &n, &m);
		for(int i=1; i<=m; i++){
			scanf("%d %d", &uu, &vv);
			add_edge(uu, vv);
		}
		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]){
					ind[bel[t]]++;
					oud[bel[i]]++;
				}
			}
		for(int i=1; i<=scc; i++){
			if(!ind[i])	cnt1++;
			if(!oud[i])	cnt2++;
		}
		if(scc==1)	printf("0
");
		else
			printf("%d
", max(cnt1, cnt2));
	}
	return 0;
}