LA4287 tarjan求強連通份量+縮點

題意:問至少還須要加入多少條有向邊才能變成強連通圖.

答案=max(in0,out0),特判scc=1.

代碼:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<stack>
 4 #define min(a,b) (a<b?a:b)
 5 #define max(a,b) (a>b?a:b)
 6 using namespace std;
 7 const int MAXE=2e5+7;
 8 const int MAXV=2e4+7;
 9 int V,E;
10 struct edge{
11     int next,v;
12 }es[MAXE];
13 int head[MAXV],tot;
14 void init(){
15     tot=0;
16     memset(head,-1,sizeof(head));
17 }
18 void addEdge(int u,int v){
19     es[tot].v=v;
20     es[tot].next=head[u];
21     head[u]=tot++;
22 }
23 int low[MAXV],dfn[MAXV],time;
24 int belong[MAXV],scc;
25 bool instack[MAXV];
26 stack<int>s;
27 void dfs(int u,int pre){
28     low[u]=dfn[u]=++time;
29     instack[u]=true;
30     s.push(u);
31     for(int i=head[u];~i;i=es[i].next){
32         int v=es[i].v;
33         if(!dfn[v]){
34             dfs(v,u);
35             low[u]=min(low[u],low[v]);
36         }else if(instack[v]){
37             low[u]=min(low[u],dfn[v]);
38         }
39     }
40     if(low[u]==dfn[u]){
41         scc++;
42         for(;;){
43             int now=s.top();s.pop();
44             belong[now]=scc;
45             instack[now]=false;
46             if(now==u)break;
47         }
48     }
49 }
50 int in[MAXV],out[MAXV];
51 int main(){
52     int T;scanf("%d",&T);
53     while(T--){
54         scanf("%d%d",&V,&E);    
55         init();
56         memset(in,0,sizeof(in));
57         memset(out,0,sizeof(out));
58         memset(low,0,sizeof(low));
59         memset(dfn,0,sizeof(dfn));
60         memset(instack,0,sizeof(instack));
61         memset(belong,0,sizeof(belong));
62         time=scc=0;
63         for(int i=0;i<E;++i){
64             int a,b;scanf("%d%d",&a,&b);
65             addEdge(a,b);
66         }
67         for(int i=1;i<=V;++i){
68             if(!dfn[i]){
69                 dfs(i,-1);
70             }
71         }
72     //    printf("scc=%d\n",scc);
73         if(scc==1){
74             printf("0\n");
75             continue;
76         }
77         for(int u=1;u<=V;++u){//縮點後的操做,對錶明元進行操做
78             for(int i=head[u];~i;i=es[i].next){
79                 int v=es[i].v;
80                 if(belong[u]!=belong[v]){
81                     out[belong[u]]++;
82                     in[belong[v]]++;
83                 }
84             }
85         }
86         int IN=0,OUT=0;
87         for(int i=1;i<=scc;++i){
88         //    printf("in[%d]=%d,out[%d]=%d\n",i,in[i],i,out[i]);
89             if(in[i]==0)++IN;
90             if(out[i]==0)++OUT;
91         }
92         printf("%d\n",max(IN,OUT));
93     }
94     return 0;
95 }

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