pku 1523

pku 1523

先求割点然后再求连同分支数量，求连同分支的时候我是把割点断开后，然后dfs求；这里容易理解，但是求割点的算法我还是看了2天。。。我太菜了，找了网上很多资料才弄明白，不容易啊！！

以下这个求连通图的模板写的简单容易明白

void DFS(int m) { dep[m] = depth; low[m] = depth; flag[m] = true; depth ++; int i; for(对于每一个从m出发的节点i){ if(没有被访问过){ DFS(i); low[m] = min(low[m],low[i]); if(low[i] >= dep[m] && m不是根节点) f[m]++; //m是割点 else if(m是根节点) son_num ++; else low[m] = min(low[m],dep[i]); } } }  

#include <iostream>#include <string>#include <vector>#define maxn 1001using namespace std;int dfn[maxn],low[maxn],vis[maxn],f[maxn],vis2[maxn];int rootson,ord;vector<int>G[maxn];int min(int a, int b){return a>b?b:a;}void add(int a, int b){vis2[a] = vis2[b] = 1;G[a].push_back(b);G[b].push_back(a);}void dfs(int n){int m,i;vis[n] = 1;dfn[n] = low[n] = ++ord;for(i = 0; i<G[n].size(); i++){m = G[n][i];if(!vis[m]){dfs(m);if(n == 1)rootson++;else{low[n] = min(low[n],low[m]);if(low[m]>=dfn[n])f[n]++;}}else low[n] = min(low[n],dfn[m]);}}void dfs1(int cur){int i;vis[cur] = 1;for(i = 0; i<G[cur].size(); i++)if(!vis[G[cur][i]])dfs1(G[cur][i]);}int main(){int i,u,v,ans,cnt = 1,p,flag;while(scanf("%d",&u)!=EOF && u!=0){memset(vis2,0,sizeof(vis2));for(i = 1; i<maxn; i++)G[i].clear();scanf("%d",&v);add(u,v);while(scanf("%d",&u),u){scanf("%d",&v);add(u,v);}memset(vis,0,sizeof(vis));memset(f,0,sizeof(f));flag = ord = rootson = ans = 0;dfs(1);if(rootson>1){f[1]++;ans++;}for(i = 2; i<maxn; i++)if(f[i])ans++;printf("Network #%d/n",cnt);cnt++;for(i = 1; i<maxn; i++){p = 0;if(f[i]){flag = 1;memset(vis,0,sizeof(vis));vis[i] = 1;for(int k = 1; k<maxn; k++)if(!vis[k]&&vis2[k]){p++;dfs(k);}printf(" SPF node %d leaves %d subnets/n",i,p);}}if(!flag)printf(" No SPF nodes/n");printf("/n");}return 0;}