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POJ——T2117 Electricity

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5459		Accepted: 1788

当m==0 时，特判输出 n-1~~~

先求出原始的连通分量，

然后只有删割点才会增加连通分量，枚举每个割点

最后加上原始的

  1 #include <algorithm>
  2 #include <cstring>
  3 #include <cstdio>
  4 
  5 using namespace std;
  6 
  7 const int N(100015);
  8 int n,m,u,v;
  9 int sumedge,head[N<<1];
 10 struct Edge
 11 {
 12     int to,next;
 13     Edge(int to=0,int next=0) :
 14         to(to),next(next) {}
 15 }edge[N*5];
 16 
 17 void ins(int from,int to)
 18 {
 19     edge[++sumedge]=Edge(to,head[from]);
 20     head[from]=sumedge;
 21 }
 22 
 23 int low[N<<1],dfn[N<<1],tim;
 24 int sumcol,num[N];
 25 int cutpoint[N];
 26 
 27 void DFS(int now,int pre)
 28 {
 29     low[now]=dfn[now]=++tim;
 30     int sumtredge=0,if_cutpoint=0;
 31     for(int i=head[now];i!=-1;i=edge[i].next)
 32     {
 33         int go=edge[i].to;
 34         if((i^1)!=pre)
 35         {
 36             if(!dfn[go])
 37             {
 38                 sumtredge++;
 39                 DFS(go,i);
 40                 if(low[go]>=dfn[now])
 41                 {
 42                     if_cutpoint=1;
 43                     num[now]++;
 44                 }
 45                 
 46                 low[now]=min(low[now],low[go]);
 47             }
 48             else low[now]=min(low[now],dfn[go]);
 49         }
 50     }
 51     if(pre==-1)
 52     {
 53         if(sumtredge>1)  cutpoint[now]=1;
 54     }
 55     else if(if_cutpoint) cutpoint[now]=1;
 56 }
 57 
 58 int ans,cnt,tmp,vis[N],root[N],cut[N];
 59 
 60 void link(int x)
 61 {
 62     vis[x]=1;
 63     for(int i=head[x];i!=-1;i=edge[i].next)
 64     {
 65         v=edge[i].to;
 66         if(!vis[v]) link(v);
 67     }
 68 }
 69 
 70 void cut_point(int pre,int fa)
 71 {
 72     vis[pre]=1;
 73     for(int i=head[pre];i!=-1;i=edge[i].next)
 74     {
 75         int go=edge[i].to;
 76         if(!vis[go])
 77         {
 78             if(pre==fa) cnt++;
 79             cut_point(go,fa);
 80         }
 81     }
 82 }
 83 
 84 void init(int n)
 85 {
 86     sumedge=-1; ans=tim=tmp=cnt=0;
 87     memset(vis,0,sizeof(vis));
 88     memset(low,0,sizeof(low));
 89     memset(dfn,0,sizeof(dfn));
 90     memset(num,0,sizeof(num));
 91     memset(cut,0,sizeof(cut));
 92     memset(root,0,sizeof(root));
 93     memset(head,-1,sizeof(head));
 94     memset(cutpoint,0,sizeof(cutpoint));
 95 }
 96 
 97 int main()
 98 {
 99 //    freopen("makerout.txt","r",stdin);
100 //    freopen("myout.txt","w",stdout);
101     
102     while(~scanf("%d%d",&n,&m)&&n)
103     {
104         init(n);
105         if(m==0) printf("%d
",n-1);
106         else
107         {
108             for(;m;m--)    
109             {
110                 scanf("%d%d",&u,&v);
111                 ins(u,v); ins(v,u);
112             }
113             for(int i=0;i<n;i++)
114                 if(!vis[i]) ++tmp,link(i);
115             for(int i=0;i<n;i++)
116                 if(!dfn[i]) DFS(i,-1);
117             for(int i=0;i<n;i++)
118                 if(cutpoint[i])
119                 {
120                     memset(vis,0,sizeof(vis));
121                     cut_point(i,i);
122                     ans=max(ans,cnt-1);
123                     cnt=0;
124                 }
125             ans+=tmp;
126             printf("%d
",ans);
127         }
128     }
129     return 0;
130 }

——每当你想要放弃的时候，就想想是为了什么才一路坚持到现在。

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原文地址：https://www.cnblogs.com/Shy-key/p/6883121.html