【BZOJ1123】【POI2008】BLO（DFS树）

Description

给定一张无向图，求每个点被封锁之后有多少个有序点对(x,y)(x!=y,1<=x,y<=n)满足x无法到达y

Solution

直接dfs乱搞嘛，用类似Tarjan的框架，求出去掉该点后每个块的大小，乘法原理算就行了。

Source

/************************************************
 * Au: Hany01
 * Date: Apr 5th, 2018
 * Prob: [BZOJ1123][POI2008] BLO
 * Email: hany01@foxmail.com
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
    register int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 100005, maxm = 1000005;

int n, e, v[maxm], nex[maxm], beg[maxn], dfn[maxn], low[maxn], tim, sz[maxn];
LL Ans[maxn];

inline void add(int uu, int vv) { v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e; }

void dfs(int u)
{
    register int t = 0;
    dfn[u] = low[u] = ++ tim, sz[u] = 1;
    for (register int i = beg[u]; i; i = nex[i])
    {
        if (!dfn[v[i]]) {
            dfs(v[i]), sz[u] += sz[v[i]];
            chkmin(low[u], low[v[i]]);
            if (low[v[i]] >= dfn[u])
                Ans[u] += (LL)sz[v[i]] * (LL)(sz[v[i]] - 1), t += sz[v[i]];
        } else if (dfn[v[i]] < dfn[u])
            chkmin(low[u], dfn[v[i]]);
    }
    Ans[u] += (LL)(n - t - 1) * (n - t - 2);
}

int main()
{
#ifdef hany01
    File("bzoj1123");
#endif

    static int m, uu, vv;

    n = read(), m = read();
    For(i, 1, m)
        uu = read(), vv = read(), add(uu, vv), add(vv, uu);

    dfs(1);

    For(i, 1, n) printf("%lld\n", (LL)n * (LL)(n - 1) - Ans[i]);

    return 0;
}
//《金陵图》
//作者：韦庄
//江雨霏霏江草齐，六朝如梦鸟空啼。
//无情最是台城柳，依旧烟笼十里堤。