poj 2186 Popular Cows (强连通分量 tarjan)

poj 2186 Popular Cows

强连通分量 tarjan

http://poj.org/problem?id=2186

 

#include <iostream> #include "stdio.h" #include "string.h" using namespace std; struct node { int v, next; }edge[50000]; const int MAXV = 10001; int adj[MAXV], lp, low[MAXV], dfn[MAXV], out[MAXV], belong[MAXV], stack[MAXV], index, top, num; bool instack[MAXV]; void addedge(int u, int v) { edge[lp].v = v; edge[lp].next = adj[u]; adj[u] = lp++; } void tarjan(int u) { low[u] = dfn[u] = ++index; stack[++top] = u; instack[u] = true; for(int i = adj[u]; i != -1; i = edge[i].next) { int v = edge[i].v; if(!dfn[v]) { tarjan(v); low[u] = min(low[u], low[v]); } else if(instack[v]) low[u] = min(low[u], dfn[v]); } int tmp; if(low[u] == dfn[u]) { num++; do { tmp = stack[top--]; instack[tmp] = false; belong[tmp] = num; }while(tmp != u); } } int main() { int nv, ne, i, j, a, b; while(scanf("%d%d", &nv, &ne) != EOF) { num = top = index = lp = 0; memset(adj, -1, sizeof(adj)); memset(dfn, 0, sizeof(dfn)); memset(out, 0, sizeof(out)); memset(instack, false, sizeof(instack)); for(i = 0; i < ne; i++) { scanf("%d%d", &a, &b); addedge(a, b); } for(i = 1; i <= nv; i++) if(!dfn[i]) tarjan(i); for(i = 1; i <= nv; i++) for(j = adj[i]; j != -1; j = edge[j].next) if(belong[i] != belong[edge[j].v]) out[belong[i]]++; int ans = 0, cnt = 0, pos; for(i = 1; i <= num; i++) if(out[i] == 0) { cnt++; //记录出度为0的强连通分量的个数 pos = i; } if(cnt == 1) { for(i = 1; i <= nv; i++) if(belong[i] == pos) ans++; } cout << ans << endl; } return 0; }