POJ 1236 Network of Schools Tarjan缩点-白红宇

POJ 1236 Network of Schools Tarjan缩点

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发布时间：2019-06-12

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Network of Schools

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22729		Accepted: 8926

Description

A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a list of schools to which it distributes software (the “receiving schools”). Note that if B is in the distribution list of school A, then A does not necessarily appear in the list of school B

You are to write a program that computes the minimal number of schools that must receive a copy of the new software in order for the software to reach all schools in the network according to the agreement (Subtask A). As a further task, we want to ensure that by sending the copy of new software to an arbitrary school, this software will reach all schools in the network. To achieve this goal we may have to extend the lists of receivers by new members. Compute the minimal number of extensions that have to be made so that whatever school we send the new software to, it will reach all other schools (Subtask B). One extension means introducing one new member into the list of receivers of one school.

Input

The first line contains an integer N: the number of schools in the network (2 <= N <= 100). The schools are identified by the first N positive integers. Each of the next N lines describes a list of receivers. The line i+1 contains the identifiers of the receivers of school i. Each list ends with a 0. An empty list contains a 0 alone in the line.

Output

Your program should write two lines to the standard output. The first line should contain one positive integer: the solution of subtask A. The second line should contain the solution of subtask B.

Sample Input

52 4 3 04 5 0001 0

Sample Output

12 哦豁，这道水题。我wa了一天，从11点WA到晚上九点。。最后发现少写一句话orz... 忘记给u标记orzorzorzorz然后发现自己上午敲的板子就是错的 wori.... 以后我只用bin神的模板！！！ 这个题很简单，还是tarjan缩点，分别找一下入度和出度为0的点。 最少通知几个，易得，考虑入度为0的强连通分量~ 最少加几条边，sum1和sum2的最大值？为什么呢，因为要让任意选一个点都满足条件，画一下图，得让每个强连通分量有入有出。所以要取最大值。 贴代码

1 #include
      
         2 #include
       
          3 #include
        
           4 #include
         
            5 #include
          
             6 #include
           
             7 #include
            
              8 #include
             
               9 #include
               10 using namespace std; 11 #define mem(a,b) memset(a,b,sizeof(a)) 12 #define ll long long 13 #define inf 1000000000 14 #define maxn 105 15 #define maxm 100005 16 struct node 17 { 18 int to,next; 19 } edge[maxm]; 20 int n,m,head[maxn],vis[maxn],dfn[maxn],low[maxn],cdu[maxn],num[maxn],cnt,timi,stack1[maxn],top,cut,rdu[maxn]; 21 stack 
               
                 s; 22 void init() 23 { 24 memset(dfn,0,sizeof(dfn)); 25 memset(low,0,sizeof(low)); 26 memset(head,-1,sizeof(head)); 27 memset(vis,0,sizeof(vis)); 28 memset(num,0,sizeof(num)); 29 memset(cdu,0,sizeof(cdu)); 30 memset(rdu,0,sizeof(rdu)); 31 while(!s.empty()) s.pop(); 32 cnt=0; 33 timi=1; 34 top=0; 35 cut=0; 36 } 37 void addedge(int u,int v) 38 { 39 edge[cnt].to=v; 40 edge[cnt].next=head[u]; 41 head[u]=cnt; 42 cnt++; 43 } 44 void tarjan(int u) 45 { 46 dfn[u]=timi; 47 low[u]=timi; 48 timi++; 49 s.push(u); 50 vis[u]=1; 51 for(int i=head[u]; i!=-1; i=edge[i].next) 52 { 53 int v=edge[i].to; 54 if(!dfn[v]) 55 { 56 tarjan(v); 57 low[u]=min(low[u],low[v]); 58 } 59 else 60 { 61 if(vis[v]) 62 low[u]=min(low[u],dfn[v]); 63 } 64 } 65 if(low[u]==dfn[u]) 66 { 67 cut++; 68 int x=s.top(); 69 while(x!=u) 70 { 71 vis[x]=0; 72 num[x]=cut; 73 s.pop(); 74 x=s.top(); 75 } 76 num[x]=cut; 77 vis[x]=0; 78 s.pop(); 79 } 80 } 81 int mp[maxn][maxn]; 82 int main() 83 { 84 int a; 85 while(cin>>n) 86 { 87 init(); 88 for(int i=1; i<=n; ++i) 89 { 90 while(~scanf("%d",&a)&&a) { 91 addedge(i,a); 92 mp[i][a]=1; 93 } 94 } 95 for(int i=1; i<=n; ++i) 96 { 97 if(!dfn[i]) tarjan(i); 98 } 99 if(cut==1)100 {101 printf("1\n0\n");102 continue;103 }104 for(int i=1; i<=n; ++i)105 for(int j=1; j<=n; ++j)106 {107 if(mp[i][j]==1&&num[i]!=num[j]) 108 {cdu[num[i]]++;rdu[num[j]]++;}109 }110 int sum1=0,sum2=0;111 for(int i=1; i<=cut; ++i)112 {113 if(!cdu[i]) sum1++;114 if(!rdu[i]) sum2++;115 }116 //cout<
                
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转载于:https://www.cnblogs.com/TYH-TYH/p/9392854.html

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