poj 1523 SPF

给一个网络， 需要求出割点和被这个割点割成的子网络的个数

#include<stdio.h>#include<string.h>const int MAXN = 1001;int low[MAXN], dfn[MAXN], ans[MAXN], map[MAXN][MAXN];int n, cnt;void dfs(int now){dfn[now] = low[now] = cnt++;for(int i = 1; i <= n; ++i){if(map[now][i]){if(dfn[i] == 0){//forward edgedfs(i);//继续递归if(low[now] > low[i])low[now] = low[i];if(dfn[now] <= low[i])ans[now]++;//直接累加} else {//back edge if(low[now] > dfn[i])low[now] = dfn[i];}}}}void init(){memset(ans, 0, sizeof(ans));memset(map, 0, sizeof(map));memset(dfn, 0, sizeof(dfn));memset(low, 0, sizeof(low));cnt = 1;}void solve(){bool ok = true;dfs(1);//1为根ans[1]--;//为根时要减一，因为为根时有大于等于两个子树时才是割点for(int i = 1; i <= n; ++i){if(ans[i] > 0){printf(" SPF node %d leaves %d subnets/n",i, ans[i] + 1);ok = false;}}if(ok){printf(" No SPF nodes/n");}printf("/n");}int main(){int a, b, cases = 0;while(scanf("%d",&a), a){init();n = 0;scanf("%d",&b);//printf("a , b =%d%d/n",a, b);map[a][b] = map[b][a] = 1;n = a > b ? a : b;while(scanf("%d",&a), a){scanf("%d",&b);//printf("a , b =%d%d/n",a, b);map[a][b] = map[b][a] = 1;n = n > (a > b ? a : b) ? n : (a > b ? a : b);}printf("Network #%d/n",++cases);solve();}return 0;}