POJ 2762 Going from u to v or from v to u? 强连通分量+DAG最长路

题意：给你一个有向图，问你是否对于任意两个点都有一条u->v或者v->u的路。

思路：对于一个强联通分量，里面的点一定满足条件，我们可以把他们缩点，而对于不在同一联通分量的点，如果存在一条路径，使得这条路径能够遍历所有点，即符合题意。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i, a, b ) for( int i = a; i < b; i++ )
#define CLR( a , x ) memset( a , x , sizeof a )

const int maxn = 1000 + 10;
const int maxe = 20000 + 10;

struct Edge{
int v , next;
Edge (int v = 0, int next = 0) : v(v), next(next) {}
};

struct SCC{
int Head[maxn], cntE;
int dfn[maxn], low[maxn], dfs_clock;
int scc[maxn], scc_cnt;
int Stack[maxn], top;
bool ins[maxn];
Edge edge[maxe];
void init(){
top = 0;
cntE = 0;
scc_cnt = 0;
dfs_clock = 0;
CLR(ins, 0);
CLR(dfn, 0);
CLR(Head, -1);
}
void add(int u, int v){
edge[cntE] = Edge( v, Head[u]);
Head[u] = cntE++;
}
void Tarjan(int u){
dfn[u] = low[u] = ++dfs_clock;
Stack[top++] = u;
ins[u] = 1;
for (int i = Head[u] ; ~i ; i = edge[i].next){
int v = edge[i].v;
if (!dfn[v]){
Tarjan (v) ;
low[u] = min(low[u], low[v]) ;
}
else if (ins[v])
low[u] = min (low[u], dfn[v]) ;
}
if (low[u] == dfn[u]){
++scc_cnt;
while ( 1 ){
int v = Stack[--top];
ins[v] = 0;
scc[v] = scc_cnt;
if (v == u)
break;
}
}
}
void find_scc(int n){
REP(i, 0, n) if(!dfn[i]) Tarjan (i) ;
}
}scc;

int n, m;
int in[maxn];
int dp[maxn];
int H[maxn], cnte;
Edge e[maxe];

void Init(){
memset(H, -1, sizeof(H));
cnte = 0;
}

void Add(int u, int v){
e[cnte] = Edge(v, H[u]);
H[u] = cnte++;
}

int DP(int u){
int &ans = dp[u];
if(ans > 0) return ans;
for(int i = H[u]; ~i; i = e[i].next)
ans = max(ans, DP(e[i].v) + 1);
return ans;
}

void solve(){
scanf("%d%d", &n, &m);
scc.init();
memset(dp, 0, sizeof(dp));
memset(in, 0, sizeof(in));
for(int i = 0; i < m; i++){
int u, v;
scanf("%d%d", &u, &v);
scc.add(u-1, v-1);
}
scc.find_scc(n);
Init();
for(int i = 0; i < n; i++)
for(int j = scc.Head[i]; ~j; j = scc.edge[j].next){
int u = i, v = scc.edge[j].v;
if(scc.scc[u] != scc.scc[v]){
Add(scc.scc[u], scc.scc[v]);
in[scc.scc[v]]++;
}
}
int i;
for(i = 1; i <= scc.scc_cnt; i++){
if(in[i] == 0){
DP(i);
break;
}
}
if(dp[i] == scc.scc_cnt - 1)
printf("Yes\n");
else
printf("No\n");
}

int main()
{
int T;
scanf("%d", &T);
while(T--) solve();
return 0;
}