Bayan 2015 Contest Warm Up E题（双连通分量缩点+DP计数）

E. Strongly Connected City 2

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Imagine a city with n junctions and m streets. Junctions are numbered from 1 to n.

In order to increase the traffic flow, mayor of the city has decided to make each street one-way. This means in the street between junctions u and v, the traffic moves only from u to v or only from v to u.

The problem is to direct the traffic flow of streets in a way that maximizes the number of pairs (u, v) where 1 ≤ u, v ≤ n and it is possible to reach junction v from u by passing the streets in their specified direction. Your task is to find out maximal possible number of such pairs.

Input

The first line of input contains integers n and m, (), denoting the number of junctions and streets of the city.

Each of the following m lines contains two integers u and v, (u ≠ v), denoting endpoints of a street in the city.

Between every two junctions there will be at most one street. It is guaranteed that before mayor decision (when all streets were two-way) it was possible to reach each junction from any other junction.

Output

Print the maximal number of pairs (u, v) such that that it is possible to reach junction v from u after directing the streets.

Sample test(s)

input

5 4
1 2
1 3
1 4
1 5

output

13

input

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

output

16

input

2 1
1 2

output

3

input

6 7
1 2
2 3
1 3
1 4
4 5
5 6
6 4

output

27

Note

In the first sample, if the mayor makes first and second streets one-way towards the junction 1 and third and fourth streets in opposite direction, there would be 13 pairs of reachable junctions: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (2, 1), (3, 1), (1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

题意：给出一个无向图，要求把每条边都变成单向的，然后使得能够到达的点对最多

思路：首先不难想到要用双连通分量缩点，缩完点以后就是一棵树了

            然后暴力以每个点（注意这里已经是缩点之后的点）为树根进行DFS，边DFS边计数即可

            还要注意一种情况，就是根的子树可以分为两部分num1 和 num2，最后的答案要累加上num1 * num2，意味着num1的所有点都可以达到num2，这个用背包求即可

            细节看代码吧~

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef long long ll;
const int MAXN = 2010;
vectorhead[MAXN],cphead[MAXN];
stacks;
int dfs_clock;
int pre[MAXN];
int id[MAXN];
int low[MAXN];
int components;
int sz[MAXN];

void tarjan(int u,int p)
{
    pre[u]=low[u]=++dfs_clock;
    s.push(u);
    int L=head[u].size();
    for(int i=0;i & q)
{
    int L=q.size();
    int to=0;
    for(int i=0;i=q[i];--j){
            f[j]|=f[j-q[i]];
        }
    }
    int ans=0;
    for(int i=1;i<=to;i++){
        if(f[i])ans=max(ans,i*(to-i));
    }
    return ans;
}

int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0;i=id[v])continue;
            cphead[id[i]].push_back(id[v]);
            cphead[id[v]].push_back(id[i]);
        }
    }
    int ans=0;
    for(int i=1;i<=components;i++){
        int L=cphead[i].size();
        tot[i]=sz[i]*sz[i];
        num[i]=sz[i];
        vectorss;
        for(int j=0;j