POJ 1523 SPF (割点）

题意：求割点，并计算去掉割点后连通分支的数量。

#include <iostream>using namespace std;#define N 1005#define min(a,b) (a<b?a:b)int size, id;int head[N], vis[N];int low[N], dfn[N], connect[N];struct { int v, next; } edge[N];void add ( int u, int v ){size++;edge[size].v = v;edge[size].next = head[u];head[u] = size;size++;edge[size].v = u;edge[size].next = head[v];head[v] = size;}void dfs ( int u ){vis[u] = true;low[u] = dfn[u] = ++id;for ( int i = head[u]; i; i = edge[i].next ){int v = edge[i].v;if ( vis[v] )low[u] = min ( low[u], dfn[v] );else{dfs ( v );low[u] = min ( low[u], low[v] );if ( low[v] >= dfn[u] ) connect[u]++;}}}int main(){int u, v, flag, i, cs = 0;while ( scanf("%d",&u) && u ){size = id = 0;memset(head,0,sizeof(head));scanf("%d",&v);add ( u, v );    while ( scanf("%d",&u) && u ){    scanf("%d",&v);    add ( u, v );}for ( i = 1; i <= 1000; i++ )vis[i] = low[i] = dfn[i] = connect[i] = 0;dfs ( 1 );connect[1]--;printf("Network #%d\n",++cs);for ( i = 1, flag = 0; i <= 1000; i++ ){if ( connect[i] <= 0 ) continue;flag = 1;printf("  SPF node %d leaves %d subnets\n", i, connect[i]+1 );}if ( flag == 0 )printf("  No SPF nodes\n");putchar('\n');}return 0;}