HDU 4635(强连通分量分解

题目:给出一个有向图,要求添加最多的边数,使得图仍然不强连通.

思路:首先这个图在添加边之后肯定变成了两个强连通分量,现在就看怎么分.然后我们可以注意到,原图进行强连通分量分解之后必然存在一些分量的出度或入度为0,最小的分量肯定在这些分量之中.那么找出这个分量就可以得出的结果了.

/*
* @author:  Cwind
*/
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <functional>
#include <set>
#include <cmath>
using namespace std;
#define IOS std::ios::sync_with_stdio (false);std::cin.tie(0)
#define pb push_back
#define PB pop_back
#define bk back()
#define fs first
#define se second
#define sq(x) (x)*(x)
#define eps (1e-7)
#define IINF (1<<29)
#define LINF (1ll<<59)
#define INF 1000000000
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> P;
inline void debug_case(){static int cas=1;printf("<---------------Case #:%d------------------>\n",cas);cas++;
}
const int maxn=1e5+300;
int T;
int n,m;
vector<int> G[maxn];
int dfn[maxn];
int clo;
int inscc[maxn],stk[maxn],low[maxn],top,sccid,cnt[maxn];
bool instk[maxn];
void tarjan(int v,int f){low[v]=dfn[v]=++clo;stk[top++]=v;instk[v]=1;for(int i=0;i<G[v].size();i++){int u=G[v][i];if(!dfn[u]){tarjan(u,v);low[v]=min(low[v],low[u]);}else if(instk[u]){low[v]=min(low[v],dfn[u]);}}if(low[v]==dfn[v]){sccid++;do{inscc[stk[--top]]=sccid;cnt[sccid]++;instk[stk[top]]=0;}while(stk[top]!=v);}
}
int d1[maxn],d2[maxn];
void init(){for(int i=0;i<=n;i++){G[i].clear();d2[i]=d1[i]=cnt[i]=dfn[i]=0;}sccid=top=clo=0;
}
int cas=0;
int main(){freopen("/home/files/CppFiles/in","r",stdin);//freopen("defense.in","r",stdin);//freopen("defense.out","w",stdout);cin>>T;while(T--){//    debug_case();scanf("%d%d",&n,&m);init();for(int i=0;i<m;i++){int a,b;scanf("%d%d",&a,&b);G[a].pb(b);}for(int i=1;i<=n;i++){if(!dfn[i]) tarjan(i,-1);}for(int i=1;i<=n;i++){for(int j=0;j<G[i].size();j++){int u=G[i][j];if(inscc[u]!=inscc[i]){d1[inscc[u]]++;d2[inscc[i]]++;}}}int minn=1e9;for(int i=1;i<=sccid;i++){if(d1[i]==0) minn=min(minn,cnt[i]);if(d2[i]==0) minn=min(minn,cnt[i]);}ll a=n-minn,b=minn;ll ans=(ll)n*(n-1)-(ll)a*b-(ll)m;if(sccid==1) ans=-1;printf("Case %d: %lld\n",++cas,ans);}return 0;
}