洛谷 P3469 [POI2008]BLO-Blockade （Tarjan，割点）

P3469 [POI2008]BLO-Blockade

https://www.luogu.org/problem/P3469

题目描述

There are exactly nn towns in Byteotia.

Some towns are connected by bidirectional roads.

There are no crossroads outside towns, though there may be bridges, tunnels and flyovers. Each pair of towns may be connected by at most one direct road. One can get from any town to any other-directly or indirectly.

Each town has exactly one citizen.

For that reason the citizens suffer from loneliness.

It turns out that each citizen would like to pay a visit to every other citizen (in his host's hometown), and do it exactly once. So exactly n\cdot (n-1)n⋅(n−1) visits should take place.

That's right, should.

Unfortunately, a general strike of programmers, who demand an emergency purchase of software, is under way.

As an act of protest, the programmers plan to block one town of Byteotia, preventing entering it, leaving it, and even passing through.

As we speak, they are debating which town to choose so that the consequences are most severe.

Task Write a programme that:

reads the Byteotian road system's description from the standard input, for each town determines, how many visits could take place if this town were not blocked by programmers, writes out the outcome to the standard output.

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

输入格式

In the first line of the standard input there are two positive integers: nn and mm (1\le n\le 100\ 0001≤n≤100 000, 1\le m\le 500\ 0001≤m≤500 000) denoting the number of towns and roads, respectively.

The towns are numbered from 1 to nn.

The following mm lines contain descriptions of the roads.

Each line contains two integers aa and bb (1\le a<b\le n1≤a<b≤n) and denotes a direct road between towns numbered aa and bb.

输出格式

Your programme should write out exactly nn integers to the standard output, one number per line. The i^{th}ith line should contain the number of visits that could not take place if the programmers blocked the town no. ii.

题意翻译

题目描述

在Byteotia有n个城镇。 一些城镇之间由无向边连接。 在城镇外没有十字路口，尽管可能有桥，隧道或者高架公路（反正不考虑这些）。每两个城镇之间至多只有一条直接连接的道路。人们可以从任意一个城镇直接或间接到达另一个城镇。 每个城镇都有一个公民，他们被孤独所困扰。事实证明，每个公民都想拜访其他所有公民一次（在主人所在的城镇）。所以，一共会有n*（n-1）次拜访。

不幸的是，一个程序员总罢工正在进行中，那些程序员迫切要求购买某个软件。

作为抗议行动，程序员们计划封锁一些城镇，阻止人们进入，离开或者路过那里。

正如我们所说，他们正在讨论选择哪些城镇会导致最严重的后果。

编写一个程序：

读入Byteotia的道路系统，对于每个被决定的城镇，如果它被封锁，有多少访问不会发生，输出结果。

输入输出格式

第一行读入n，m，分别是城镇数目和道路数目

城镇编号1~n

接下来m行每行两个数字a,b，表示a和b之间有有一条无向边

输出n行，每行一个数字，为第i个城镇被锁时不能发生的访问的数量。

翻译提供者：Park

输入输出样例

输入 #1复制

5 5
1 2
2 3
1 3
3 4
4 5

输出 #1复制

8
8
16
14
8

思路：

每一个点被封锁后，一定会有$ 2*(n-1)$ 个城市对无法到达，即该城市到剩余\(n-1\) 个城市和剩余$ n-1 $ 个城市到该城市。

还有一种情况，即一个其他城市对\(u->v\) 必须经过点x，那么节点x被封锁后，也应该计算 u->v 这个贡献。

在学习Tarjan算法的过程中我们知道无向图中的割点就是把这个节点去掉后，剩下的图变得不联通。

来看下图中的2号节点，如果该节点封锁，那么上面的“0,1,4,5” 节点就无法和下面的“3”节点连接。

那么每一个节点封锁后无法相互到达的节点对就要加上：

封锁后产生的连通块中的节点个数相互乘起来*2

封锁后产生的连通块中的节点个数可以在Tarjan算法中维护出来。

注意要把Tarjan中的

if (v == pre) { continue; }

这个语句要去掉，不然无法维护出Tarjan第一个访问的节点的答案，

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}

inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
const int MAXN = 100010;
const int MAXM = 1000010;
struct Edge {
    int to, next;
    bool cut;//是否为桥的标记
} edge[MAXM];
int head[MAXN], tot;
int Low[MAXN], DFN[MAXN], Stack[MAXN];
int Index, top;
bool Instack[MAXN];
bool cut[MAXN];
int add_block[MAXN];//删除一个点后增加的连通块
int n, m;
void addedge(int u, int v)
{
    edge[tot].to = v; edge[tot].next = head[u]; edge[tot].cut = false;
    head[u] = tot++;
}
ll cntson[MAXN];
ll vans[maxn];
void Tarjan(int u, int pre)
{
    ll z = 0ll;
    int v;
    cntson[u] = 1ll;
    Low[u] = DFN[u] = ++Index;
    for (int i = head[u]; i != -1; i = edge[i].next) {
        v = edge[i].to;
        if ( !DFN[v] ) {
            Tarjan(v, u);
            cntson[u] += cntson[v];
            if (Low[u] > Low[v]) {
                Low[u] = Low[v];
            }
            if (Low[v] >= DFN[u]) { //不是树根
                vans[u] += z * cntson[v];
                z += cntson[v];
            }
        } else if ( Low[u] > DFN[v]) {
            Low[u] = DFN[v];
        }
    }
    vans[u] += z * (n - z - 1ll);
}
void solve(int N)
{
    Index = top = 0;
    for (int i = 1; i <= N; i++)
        if (!DFN[i]) {
            Tarjan(i, i);
        }
    ll ans = 0ll;
    repd(i, 1, N) {
        printf("%lld\n", vans[i] + n - 1 << 1 );
    }
}
void init()
{
    tot = 0;
    memset(head, -1, sizeof(head));
}
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    init();
    scanf("%d %d", &n, &m);
    repd(i, 1, m) {
        int u, v;
        du2(u, v);
        addedge(u, v);
        addedge(v, u);
    }
    solve(n);
    return 0;
}

inline void getInt(int *p)
{
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    } else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}