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;}