|poj 2186|强连通分量|Popular Cows

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poj 2186

这题求所有满足被所有点能够到达的节点，那么我们可以进行缩点，缩点之后得到一个有向DAG图，统计新图的出度，如果有一个强连通分量的出度是=0的，那么输出这个强连通分量的大小，如果有多个，输出0

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<stack>#define ms(i ,j) memset(i, j, sizeof i)#define rd2(a, b) scanf("%d%d", &a, &b)#define rd3(a, b, c) scanf("%d%d%d", &a, &b, &c)#define FN2 "poj2186"using namespace std;const int MAXN = 10000 + 5;int n,m;vector<int> G[MAXN];int cnt, scc_size[MAXN], scc_out[MAXN], scc_belong[MAXN];int tb, dn[MAXN], low[MAXN], ex[MAXN];stack<int> s;void tarjan(int u){    dn[u] = low[u] = ++tb;    ex[u] = -1; s.push(u);    for (int i=0;i<G[u].size();i++)    {        int v = G[u][i];        if (ex[v]==0)        {            tarjan(v);            low[u] = min(low[u], low[v]);        } else if (ex[v]==-1)        {            low[u] = min(low[u], dn[v]);        }    }    if (dn[u]==low[u])        {            cnt++;            int e;            do            {                e = s.top(); s.pop();                scc_belong[e] = cnt;                scc_size[cnt]++;                ex[e] = 1;            }while (e!=u);        }}void rebuild(){    for (int x=1;x<=n;x++)    {        for (int i=0;i<G[x].size();i++)        {            int v = G[x][i];            if (scc_belong[x]!=scc_belong[v])            {                scc_out[scc_belong[x]]++;                break;            }        }    }}void init(){    for (int i=1;i<=m;i++)    {        int a,b; rd2(a, b);        G[a].push_back(b);    }    cnt = tb = 0; ms(ex, 0); ms(scc_size, 0); ms(scc_out, 0);}void solve(){    for (int i=1;i<=n;i++)    if (!dn[i]) tarjan(i);    rebuild();    int tot = 0, ans = 0;    for (int i=1;i<=cnt;i++)    if (scc_out[i]==0)    {        tot++; ans = scc_size[i];        if (tot>=2)         {            ans = 0;            break;        }     }    printf("%d\n", ans);}int main(){    while (rd2(n,m)==2)    {        init();        solve();    }    fclose(stdin); fclose(stdout);    return 0;}

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