[POI2008]BLO-Blockade【Tarjan缩点 圆方树】

题目链接

  有N个点，M条边组成的无向图，现在我们想知道删除一个点u，它会使得哪些x、y对，不能从x到达y，求对于每个点这样的<x,y>对的个数。

  于是，很明显的就是，我们可以将它缩点成为一个广义圆方树，然后在树上跑就是了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define pii pair<int, int>
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7, maxM = 1e6 + 7;
int N, M, head[maxN], cnt;
struct Eddge
{
    int nex, to;
    Eddge(int a=-1, int b=0):nex(a), to(b) {}
} edge[maxM];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v);
    head[u] = cnt++;
}
inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }
int dfn[maxN], low[maxN], tot, Stap[maxN], Stop;
int n;
bool iscut[maxN] = {false};
vector<int> V, to[maxN << 1];
void Tarjan(int u, int fa)
{
    int child = 0;
    V.push_back(u);
    dfn[u] = low[u] = ++tot;
    Stap[++Stop] = u;
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == fa) continue;
        if(!dfn[v])
        {
            child++;
            Tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if(low[v] >= dfn[u])
            {
                iscut[u] = true;
                if(low[v] == dfn[u])
                {
                    int p; ++n;
                    do
                    {
                        p = Stap[Stop--];
                        to[p].push_back(n);
                        to[n].push_back(p);
                    } while(p ^ v);
                    to[u].push_back(n);
                    to[n].push_back(u);
                }
                else
                {
                    Stop--;
                    to[u].push_back(v);
                    to[v].push_back(u);
                }
            }
        }
        else low[u] = min(low[u], dfn[v]);
    }
    if(!fa && child <= 1) iscut[u] = false;
}
ll ans[maxN] = {0}, siz[maxN << 1], all;
void dfs(int u, int fa)
{
    siz[u] = u <= N;
    for(int v : to[u])
    {
        if(v == fa) continue;
        dfs(v, u);
        siz[u] += siz[v];
    }
    if(u <= N)
    {
        ans[u] = 2LL * (all - 1);
        if(iscut[u])
        {
            for(int v : to[u])
            {
                if(v == fa)
                {
                    ans[u] += (all - siz[u]) * (siz[u] - 1LL);
                }
                else
                {
                    ans[u] += siz[v] * (all - siz[v] - 1);
                }
            }
        }
    }
}
inline void init()
{
    cnt = 0;
    for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
    scanf("%d%d", &N, &M); n = N;
    init();
    for(int i=1, u, v; i<=M; i++)
    {
        scanf("%d%d", &u, &v);
        _add(u, v);
    }
    for(int i=1; i<=N; i++)
    {
        if(!dfn[i])
        {
            V.clear();
            Tarjan(i, 0);
            all = V.size();
            dfs(i, 0);
        }
    }
    for(int i=1; i<=N; i++) printf("%lld\n", ans[i]);
    return 0;
}