hdu 4587 TWO NODES 暴力枚举+tarjan

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4587

 

题意是拿掉两个点

求最多可以把整个图分成几个联通块

 

注意到有一个模板是可以通过找割点来快速求出

“删一个点最多可以把整个图分成几个联通块”

所以这个时候要观察到点数只有5000

要大胆暴力枚举另一个点

即

先枚举一个点，然后另一个点套用tarjan模板

即可求出答案

 

#include <cstring>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <set>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;

const int maxn = 5010;
const int maxm = 10010;

struct Edge
{
    int to, next;
    bool cut;
}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];
int bridge;
bool flag[maxn];

void addedge(int u, int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    edge[tot].cut = false;
    head[u] = tot++;
}

void Tarjan(int u, int pre)
{
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    int son = 0;
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        v = edge[i].to;
        if(v == pre || flag[v] == true)
            continue;
        if(!DFN[v])
        {
            son++;
            Tarjan(v, u);
            if(Low[u] > Low[v])
                Low[u] = Low[v];

            if(Low[v] > DFN[u])
            {
                bridge++;
                edge[i].cut = true;
                edge[i^1].cut = true;
            }

            if(u != pre && Low[v] >= DFN[u])
            {
                cut[u] = true;
                add_block[u]++;
            }
        }
        else if(Low[u] > DFN[v])
            Low[u] = DFN[v];
    }

    if(u == pre & son > 1)
        cut[u] = true;
    if(u == pre)
        add_block[u] = son - 1;
    Instack[u] = false;
    top--;
}

void solve(int N)
{
    int ans = 0;
    for(int k = 1; k <= N; k++)
    {
        flag[k] = true;

        memset(DFN, 0, sizeof(DFN));
        memset(Instack, 0, sizeof(Instack));
        memset(add_block, 0, sizeof(add_block));
        memset(cut, false, sizeof(cut));
        for(int i = 0; i < tot; i++)
            edge[i].cut = false;

        Index = top = 0;
        int cnt = 0;
        for(int i = 1; i <= N; i++)
        {
            if(flag[i])
                continue;

            if(!DFN[i])
            {
                Tarjan(i, i);
                cnt++;
            }
        }
        int anss = 0;
        for(int i = 1; i <= N; i++)
            if(flag[i] == false)
                anss = max(anss, cnt + add_block[i]);

       // if(N-1 == cnt)
           // ans--;

        ans = max(ans, anss);

        flag[k] = false;
    }

    printf("%d\n", ans);
}

void init()
{
    tot = 0;
    memset(head, -1, sizeof(head));
}

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);

    int n, m;

    while(scanf("%d%d", &n, &m) == 2)
    {
        init();
        for(int i = 0; i < m; i++)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            u++;
            v++;
            addedge(u, v);
            addedge(v, u);
        }

        solve(n);
    }

    return 0;
}