BZOJ 1123: [POI2008]BLO

1123: [POI2008]BLO

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 1030  Solved: 440
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Description

Byteotia城市有n个 towns m条双向roads. 每条 road 连接 两个不同的 towns ,没有重复的road. 所有towns连通。

Input

输入n<=100000 m<=500000及m条边

Output

输出n个数，代表如果把第i个点去掉，将有多少对点不能互通。

Sample Input

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

Sample Output

8
8
16
14
8

HINT

 

Source

 

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分析

如果一个点不是割点，那么删去后不会对其他点之间的连通性造成影响；如果是一个割点，影响只和其连接的几个块的大小有关。因此Tarjan求割点的同时注意维护子树大小即可。

代码

#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

#define N 5000000
#define LL long long

int n, m; LL ans[N];

int hd[N], to[N], nt[N], tot;

int dfn[N], low[N], tim;

int tarjan(int u, int f)
{   
    dfn[u] = low[u] = ++tim; ans[u] = (n - 1) << 1;
    
    int cnt = 0, siz; LL sum = 0, tmp = 0;
    
    for (int i = hd[u]; ~i; i = nt[i])if (f != to[i])
    {
        if (!dfn[to[i]])
        {
            siz = tarjan(to[i], u);
            low[u] = min(low[u], low[to[i]]);
            if (low[to[i]] >= dfn[u])
                ans[u] += 2LL * siz * sum, sum += siz;
            else
                tmp += siz;
        }
        else
            low[u] = min(low[u], dfn[to[i]]);
    }
    
    ans[u] += 2LL * (n - (sum + 1)) * sum;
    
    return sum + tmp + 1;
}

signed main(void)
{
    scanf("%d%d", &n, &m);
    
    memset(hd, -1, sizeof(hd)), tot = 0;
    
    for (int i = 1; i <= m; ++i)
    {
        int x, y; scanf("%d%d", &x, &y);
        
        nt[tot] = hd[x]; to[tot] = y; hd[x] = tot++;
        nt[tot] = hd[y]; to[tot] = x; hd[y] = tot++;
    }
    
    memset(dfn, 0, sizeof(dfn)); tim = 0; tarjan(1, -1);
    
    for (int i = 1; i <= n; ++i)
        printf("%lld
", ans[i]);
}

#include <cstdio>

template <class T>
inline T min(const T &a, const T &b)
{
    return a < b ? a : b;
}

typedef long long lnt;

const int mxn = 100005;
const int mxm = 1000005;

int n, m;

int hd[mxn];
int to[mxm];
int nt[mxm];

int dfn[mxn];
int low[mxn];

lnt ans[mxn];

lnt tarjan(int u, int f)
{
    static int tim = 0;

    ans[u] = (n - 1) << 1;
    dfn[u] = low[u] = ++tim;

    lnt siz, sum = 0, tmp = 0;

    for (int i = hd[u], v; i; i = nt[i])
        if ((v = to[i]) != f)
        {
            if (!dfn[v])
            {
                siz = tarjan(v, u);

                low[u] = min(low[u], low[v]);

                if (low[v] >= dfn[u])
                    ans[u] += 2LL * sum * siz, sum += siz;
                else
                    tmp += siz;
            }
            else
                low[u] = min(low[u], dfn[v]);
        }

    ans[u] += 2LL * (n - sum - 1) * sum;

    return sum + tmp + 1;
}

signed main(void)
{
    scanf("%d%d", &n, &m);

    for (int i = 0; i < m; ++i)
    {
        static int x, y, tot;

        scanf("%d%d", &x, &y);

        nt[++tot] = hd[x], to[tot] = y, hd[x] = tot;
        nt[++tot] = hd[y], to[tot] = x, hd[y] = tot;
    }

    tarjan(1, 0);

    for (int i = 1; i <= n; ++i)
        printf("%lld
", ans[i]);
}

@Author: YouSiki