Hdu 4635 Strongly CTED (Tarjan)

After reading the report, the idea is the same. I just need to paste it.

After adding an edge, the graph can be divided into two parts X and Y. Only the side from X to Y has no edge from Y to X, so we need to make the number of sides as much as possible, the X part must be a complete graph, and the Y part is also. At the same time, each vertex in the X part has an edge. Assume that the X part has x points, Y has y points, x + y = n, and the number of edge F = x * y + x * (x-1) + y * (Y-1), sorted out: F = N * N-x * y. When x + y is set, the closer the two are, the greater x * y, the maximum number of edges, therefore, the gap between the number of points in Part X and part Y will be larger. Therefore, first, for the given digraph contraction point, for each point after the contraction point, if its degree of exit or entry is 0, it may become X or Y. Therefore, only the exit degree after the contraction or the entry degree is 0, the vertex that contains the least number of nodes makes it a part. All other vertices add up to another part to get the graph with the maximum number of edges.

 

 

//109MS 2916KB#include <stdio.h>#include <string.h>#define LL long longconst 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;}