BZOJ 2208 连通数(强连通分量)

先缩点，对于缩完点后的DAG，可以直接在每个scc dfs一次就可以求出终点是这个scc的点的点对个数。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi 3.1415926535

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int res=, flag=;

    char ch;

    if((ch=getchar())=='-') flag=;

    else if(ch>=''&&ch<='') res=ch-'';

    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');

    return flag?-res:res;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

struct Edge{int p, next;}edge[N*N*];

int head[N], cnt=;

char s[N][N];

int G[N][N], n;

int Low[N], DFN[N], Stack[N], Belong[N], Index, top, scc, num[N];

bool Instack[N], vis[N][N], mark[N];

LL siz[N];

void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;}

void Tarjan(int u){

    Low[u]=DFN[u]=++Index; Stack[top++]=u; Instack[u]=true;

    FOR(v,,n) {

        if (!G[u][v]) continue;

        if (!DFN[v]) {

            Tarjan(v);

            if (Low[u]>Low[v]) Low[u]=Low[v];

        }

        else if (Instack[v]&&Low[u]>DFN[v]) Low[u]=DFN[v];

    }

    int v;

    if (Low[u]==DFN[u]) {

        scc++;

        do{

            v=Stack[--top]; Instack[v]=false;

            Belong[v]=scc; num[scc]++;

        }while (v!=u);

    }

}

void solve(int nn){

    mem(DFN,); mem(Instack,); mem(num,);

    Index=scc=top=;

    FOR(i,,nn) if (!DFN[i]) Tarjan(i);

}

void dfs(int x, int fa){

    siz[x]+=num[x]*num[fa];

    mark[x]=;

    for (int i=head[x]; i; i=edge[i].next) {

        int v=edge[i].p;

        if (mark[v]) continue;

        dfs(v,fa);

    }

}

int main ()

{

    LL ans=;

    scanf("%d",&n);

    FOR(i,,n) scanf("%s",s[i]+);

    FOR(i,,n) FOR(j,,n) G[i][j]=s[i][j]-'';

    solve(n);

    FOR(i,,n) FOR(j,,n) {

        if (!G[i][j]) continue;

        int u=Belong[i], v=Belong[j];

        if (u==v||vis[u][v]) continue;

        add_edge(u,v); vis[u][v]=true;

    }

    FOR(i,,scc) mem(mark,), dfs(i,i);

    FOR(i,,scc) ans+=siz[i];

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

    return ;

}