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-9
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=2005;
//Code begin...

struct Edge{int p, next;}edge[N*N*2];
int head[N], cnt=1;
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,1,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,0); mem(Instack,0); mem(num,0);
    Index=scc=top=0;
    FOR(i,1,nn) if (!DFN[i]) Tarjan(i);
}
void dfs(int x, int fa){
    siz[x]+=num[x]*num[fa];
    mark[x]=1;
    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=0;
    scanf("%d",&n);
    FOR(i,1,n) scanf("%s",s[i]+1);
    FOR(i,1,n) FOR(j,1,n) G[i][j]=s[i][j]-'0';
    solve(n);
    FOR(i,1,n) FOR(j,1,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,1,scc) mem(mark,0), dfs(i,i);
    FOR(i,1,scc) ans+=siz[i];
    printf("%lld\n",ans);
    return 0;
}