ZOJ1119(SPF)

题目链接:传送门

题目大意:一副无向图，问有多少个节点满足删除该节点后图不连通，对于每个满足条件的节点，输出节点编号及删除节点将图分为几个连通块。若没有节点满足则输出No SPF nodes

题目思路:tarjan算法求关节点入门题(也可用矩阵存图)

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <algorithm>
 #include <cstring>
 #include <stack>
 #include <cctype>
 #include <queue>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <climits>
 #define lson root<<1,l,mid
 #define rson root<<1|1,mid+1,r
 #define fi first
 #define se second
 #define ping(x,y) ((x-y)*(x-y))
 #define mst(x,y) memset(x,y,sizeof(x))
 #define mcp(x,y) memcpy(x,y,sizeof(y))
 #define Min(x,y) (x<y?x:y)
 #define Max(x,y) (x>y?x:y)
 using namespace std;
 #define gamma 0.5772156649015328606065120
 #define MOD 100000007
 #define inf 0x3f3f3f3f
 #define N 1000005
 #define maxn 1000050
 typedef long long LL;
 typedef pair<int,int> PII;

 int n,m,head[],hcnt,son,num;
 int dfn[],low[],area[],vis[];
 int deep;
 struct Node{
     int to,next;
     Node(){}
     Node(int a,int b):to(a),next(b){}
 }node[];
 inline void add(int x,int y){
     node[hcnt]=Node(y,head[x]);
     head[x]=hcnt++;
 }

 void init(){
     num=;  ///最大节点的编号
     hcnt=;
     deep=; ///深度搜索树的深度优先数(第几个被搜索)
     son=;  ///记录根节点的子女个数
     low[]=dfn[]=;  ///low表示节点能通过子女和回边到达的最小深度优先数
     mst(head,-);     ///dfn表示节点的深度优先数
     mst(vis,);
     mst(area,);  ///记录删除某个节点后图被分成了几块
     vis[]=; ///标记节点是否访问过
 }

 void dfs(int x){
     for(int i=head[x];~i;i=node[i].next){
         int e=node[i].to;
         if(vis[e]) low[x]=Min(low[x],dfn[e]);
         else{
             vis[e]=;
             dfn[e]=low[e]=++deep;
             dfs(e);
             low[x]=Min(low[x],low[e]);
             if(low[e]>=dfn[x]){
                 if(x==) ++son;
                 else ++area[x];
             }
         }
     }
 }

 int main(){
     int i,j,group,Case=,x,y;
     while(scanf("%d",&x)!=EOF&&x){
         init();
         num=Max(num,x);
         scanf("%d",&y);
         num=Max(num,y);
         add(x,y);
         add(y,x);
         while(scanf("%d",&x)&&x){
             num=Max(num,x);
             scanf("%d",&y);
             num=Max(num,y);
             add(x,y);
             add(y,x);
         }
         if(Case)printf("\n");
         printf("Network #%d\n",++Case);
         dfs();
         if(son>)area[]=son-;
         int flag=;
         for(i=;i<=num;++i)if(area[i]){
             flag=;
             printf("  SPF node %d leaves %d subnets\n",i,area[i]+);
         }
         if(!flag) printf("  No SPF nodes\n");
     }
     return ;
 }