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;
}