題目鏈接:
http://acm.hdu.edu.cn/showproblem.php?pid=4635
Strongly connected
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1568 Accepted Submission(s): 654
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.
題目意思:
給一幅圖,求最多增加多少邊,可以使該圖能夠不滿足強連通性。
解題思路:
分析知,最後肯定是兩個完全圖a,b,然後a中每個節點到b中每個節點都有一條有向邊。整體考慮總的邊數爲a*(a-1)+b*(b-1)+a*b=n*n-n-a*b.
ans=n*n-n-a*b-m n/m已知,所以讓ans最大,就需讓a*b最小,使得a,b差距最大。所以找最小的a或b.
tarjan求強連通,然後求出入度或出度爲0的強連通分量,統計這些聯通分量中節點數最小的a.
代碼:
//#include<CSpreadSheet.h>
#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define Maxn 110000
int low[Maxn],dfn[Maxn],sta[Maxn],bc,sc,dep;
ll n,m;
int in[Maxn],dei[Maxn],deo[Maxn],nu[Maxn];
vector<vector<int> >myv;
bool iss[Maxn];
void tarjan(int cur)
{
int ne;
low[cur]=dfn[cur]=++dep;
sta[++sc]=cur;
iss[cur]=true;
for(int i=0;i<myv[cur].size();i++)
{
ne=myv[cur][i];
if(!dfn[ne])
{
tarjan(ne);
if(low[ne]<low[cur])
low[cur]=low[ne];
}
else if(iss[ne]&&dfn[ne]<low[cur])
low[cur]=dfn[ne];
}
if(low[cur]==dfn[cur])
{
bc++;
do
{
ne=sta[sc--];
iss[ne]=false;
in[ne]=bc;
nu[bc]++;
}while(ne!=cur);
}
}
void solve()
{
dep=sc=bc=0;
memset(nu,0,sizeof(nu));
memset(iss,false,sizeof(iss));
memset(dfn,0,sizeof(dfn));
for(int i=1;i<=n;i++)
if(!dfn[i])
tarjan(i);
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int t,kcas=0;
scanf("%d",&t);
while(t--)
{
scanf("%I64d%I64d",&n,&m);
myv.clear();
myv.resize(n+1);
for(int i=1;i<=m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
myv[a].push_back(b);
}
solve();
if(bc==1)
{
printf("Case %d: -1\n",++kcas);
continue;
}
memset(dei,0,sizeof(dei));
memset(deo,0,sizeof(deo));
for(int i=1;i<=n;i++)
for(int j=0;j<myv[i].size();j++)
{
int ne=myv[i][j];
if(in[ne]!=in[i])
{
dei[in[ne]]++;
deo[in[i]]++;
}
}
ll ans=INF;
for(int i=1;i<=bc;i++)
{
if(!dei[i])
ans=min(ans,(ll)nu[i]);
else if(!deo[i])
ans=min(ans,(ll)nu[i]);
}
printf("Case %d: %I64d\n",++kcas,n*n-n-ans*(n-ans)-m);
}
return 0;
}