Strongly connected-HDU4635
Problem - 4635 http://acm.hdu.edu.cn/showproblem.php?pid=4635
题目大意:
n个点,m条边,求最多再加几条边,然后这个图不是强连通
分析:
这是一个单向图,如果强连通的话,他最多应该有n*(n-1)条边,假设有a个强连通块,任取其中一个强连通块,假设取出的这个强连通块里有x个点,剩下的(n-a)个点看成一个强连通块,如果让这两个强连通块之间不联通,肯定是这两个只有一个方向的边,最多就会有x*(n-x)条边 所以最多加n*(n-1)-x*x(n-x)-m边。所以当x最小是式子最大。
A simple directed graph is a directed graph having no multiple edges or graph loops.
A strongly connected digraph is a directed graph in which it is possible to reach any node starting from any other node by traversing edges in the direction(s) in which they point.
Then T cases follow, each case starts of two numbers N and M, 1<=N<=100000, 1<=M<=100000, representing the number of nodes and the number of edges, then M lines follow. Each line contains two integers x and y, means that there is a edge from x to y.
If the original graph is strongly connected, just output -1.
3 3
1 2
2 3
3 1
3 3
1 2
2 3
1 3
6 6
1 2
2 3
3 1
4 5
5 6
6 4
Case 2: 1
Case 3: 15
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector> using namespace std;
#define N 100005
#define INF 0x3f3f3f3f struct node
{
int to,next;
}edge[N*]; int low[N],dfn[N],Time,top,ans,Stack[N],belong[N],sum,head[N],aa[N],in[N],out[N],Is[N]; void Inn()
{
memset(low,,sizeof(low));
memset(dfn,,sizeof(dfn));
memset(Stack,,sizeof(Stack));
memset(belong,,sizeof(belong));
memset(head,-,sizeof(head));
memset(aa,,sizeof(aa));
memset(in,,sizeof(in));
memset(out,,sizeof(out));
memset(Is,,sizeof(Is));
Time=top=ans=sum=;
} void add(int from,int to)
{
edge[ans].to=to;
edge[ans].next=head[from];
head[from]=ans++;
} void Tarjin(int u,int f)
{
low[u]=dfn[u]=++Time;
Stack[top++]=u;
Is[u]=;
int v;
for(int i=head[u];i!=-;i=edge[i].next)
{
v=edge[i].to;
if(!dfn[v])
{
Tarjin(v,u);
low[u]=min(low[u],low[v]);
}
else if(Is[v])
low[u]=min(low[u],dfn[v]);
}
if(dfn[u]==low[u])
{
sum++;
do
{
v=Stack[--top];
belong[v]=sum;
aa[sum]++;
Is[v]=;
}while(v!=u);
}
} void solve(int n,int m)
{
for(int i=;i<=n;i++)
{
if(!dfn[i])
Tarjin(i,);
}
if(sum==)
{
printf("-1\n");
return ;
}
long long Max=;
for(int i=;i<=n;i++)
{
for(int j=head[i];j!=-;j=edge[j].next)
{
int u=belong[i];
int v=belong[edge[j].to];
if(u!=v)
{
in[v]++;
out[u]++;
}
}
}
long long c=n*(n-)-m;
for(int i=;i<=sum;i++)
{
if(!in[i] || !out[i])
Max=max(Max,c-(aa[i]*(n-aa[i])));
}
printf("%lld\n",Max);
}
int main()
{
int T,n,m,a,b,i,t=;
scanf("%d",&T);
while(T--)
{
Inn();
scanf("%d %d",&n,&m);
for(i=;i<m;i++)
{
scanf("%d %d",&a,&b);
add(a,b);
}
printf("Case %d: ",t++);
solve(n,m);
}
return ;
}
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