hdu 4635 Strongly connected(Tarjan)

做完后，看了解题报告，思路是一样的。我就直接粘过来吧

最终添加完边的图，肯定可以分成两个部X和Y，其中只有X到Y的边没有Y到X的边，那么要使得边数尽可能的多，则X部肯定是一个完全图，Y部也是，同时X部中每个点到Y部的每个点都有一条边，假设X部有x个点，Y部有y个点，有x+y=n，同时边数F=x*y+x*(x-1)+y*(y-1)，整理得：F=N*N-N-x*y，当x+y为定值时，二者越接近，x*y越大，所以要使得边数最多，那么X部和Y部的点数的个数差距就要越大，所以首先对于给定的有向图缩点，对于缩点后的每个点，如果它的出度或者入度为0，那么它才有可能成为X部或者Y部，所以只要求缩点之后的出度或者入度为0的点中，包含节点数最少的那个点，令它为一个部，其它所有点加起来做另一个部，就可以得到最多边数的图了

 

//109MS 	2916KB

#include <stdio.h>

#include <string.h>

#define LL long long

const int M = 100005;

const int inf = 0x3f3f3f3f;

struct Edge

{

    int to,nxt;

} edge[M];



int head[M],low[M],dfn[M],stack[M+10];

int vis[M],out[M],in[M],belong[M];

int scc,cnt ,top,ep;

LL n,m;

int min (int a,int b)

{

    return a > b ? b : a;

}

void addedge (int cu,int cv)

{

    edge[ep].to = cv;

    edge[ep].nxt = head[cu];

    head[cu] = ep ++;

}



void Tarjan(int u)

{

    int v;

    dfn[u] = low[u] = ++cnt;

    stack[top++] = u;

    vis[u] = 1;

    for (int i = head[u]; i != -1; i = edge[i].nxt)

    {

        v = edge[i].to;

        if (!dfn[v])

        {

            Tarjan(v);

            low[u] = min(low[u],low[v]);

        }

        else if (vis[v]) low[u] = min(low[u],dfn[v]);

    }

    if (dfn[u] == low[u])

    {

        ++scc;

        do

        {

            v = stack[--top];

            vis[v] = 0;

            belong[v] = scc;

        }

        while (u != v);

    }

}



void solve()

{

    int u,v;

    scc = top = cnt = 0;

    memset (vis,0,sizeof(vis));

    memset (dfn,0,sizeof(dfn));

    memset (out,0,sizeof(out));

    memset (in,0,sizeof(in));

    for (u = 1; u <= n; u ++)

        if (!dfn[u])

            Tarjan(u);



    for (u = 1; u <= n; u ++)

    {

        for (int i = head[u]; i != -1; i =edge[i].nxt)

        {

            v = edge[i].to;

            if (belong[u] != belong[v])

            {

                out[belong[u]] ++;

                in[belong[v]] ++;

            }

        }

    }

    int num[M],Min;

    memset (num,0,sizeof(num));

    for (u = 1; u <= n; u ++)

        if (!in[belong[u]]) num[belong[u]] ++;

    for (u = 1; u <= scc; u ++)

        if (num[u]!= 0&&num[u]<Min)

            Min = num[u];



    memset (num,0,sizeof(num));

    for (u = 1; u <= n; u ++)

        if (!out[belong[u]]) num[belong[u]] ++;

    for (u = 1; u <= scc; u ++)

        if (num[u]!= 0&&num[u]<Min)

            Min = num[u];

    if (scc == 1)

    {

        printf ("-1\n");

        return ;

    }

    LL ans = n*(n-1)-Min*(n-Min) - m;

    printf ("%I64d\n",ans);

}

int main ()

{

#ifdef LOCAL

    freopen("in.txt","r",stdin);

#endif

    int T,u,v,cnt = 0;

    scanf ("%d",&T);

    while (T --)

    {

        scanf ("%I64d%I64d",&n,&m);

        ep = 0;

        memset (head,-1,sizeof(head));

        for (int i = 0; i < m; i++)

        {

            scanf ("%d%d",&u,&v);

            addedge(u,v);

        }

        printf ("Case %d: ",++cnt);

        solve();

    }

    return 0;

}