bzoj2730

tarjan求割点

我发现我还不会求割点

首先我们发现如果整个图是一个点双，那么要放两个出口。答案是2 c(n, 2)

如果不是，说明这个图存在割点能把图分成很多个部分，那么我们就要把割点求出来，每个点双和割点缩成一个点，这样就构成了一棵树。然后每个度数为一的点都要放一个出口，如果度数大于一就不用放，因为怎么割都和另一边的出口相连。

求割点还是用tarjan，如果一个点的dfn<=low[son]，说明儿子没有返祖边，那么这个点去掉肯定使儿子的那个连通块独立，所以这个点就是割点，标记一下，然后dfs染其他非割点的点，然后连边算度数算大小。

奥妙重重的是low[u]=min(low[u],low[v])会错，这也告诉我们还是要按照标准来写。

upd:因为可能v本身是个割点，如果翻上去就无法发现v是割点了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 510;
int n, m, all, cnt, top, kase;
int dfn[N], low[N], vis[N], belong[N], d[N], size[N];
vector<int> G[N], g[N << 1];
void tarjan(int u, int last)
{
    dfn[u] = low[u] = ++all; 
    int child = 0;
    for(int i = 0; i < G[u].size(); ++i)
    {
        int v = G[u][i];
        if(!dfn[v]) 
        {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            ++child;
            if((u != 1 && low[v] >= dfn[u]) || (u == 1 && child > 1) && !belong[u])
            {    
                vis[u] = 1;
                belong[u] = ++cnt;
                size[cnt] = 1;
            }
        }
        else if(v != last)
            low[u] = min(low[u], dfn[v]);
    }
}
void dfs(int u)
{
    vis[u] = 1; 
    belong[u] = cnt;
    ++size[cnt];
    for(int i = 0; i < G[u].size(); ++i)
    {
        int v = G[u][i];
        if(!vis[v]) 
            dfs(v);
    }    
}
int main()
{
//    freopen("bzoj_2730.in", "r", stdin);
//    freopen("bzoj_2730.out", "w", stdout);
    while(scanf("%d", &n))
    {
        if(!n) break;
        for(int i = 1; i <= n; ++i)
        {
            int u, v; scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
            m = max(m, max(u, v));
        }
        tarjan(1, 0);
        for(int i = 1; i <= m; ++i) if(!vis[i])
        {
            ++cnt;
            dfs(i);
        }
        for(int i = 1; i <= m; ++i) 
            for(int j = 0; j < G[i].size(); ++j)
            {
                int v = G[i][j];
                if(belong[i] != belong[v])
                {
                    g[belong[i]].push_back(belong[v]);
                    g[belong[v]].push_back(belong[i]);
                }
            }
        if(cnt == 1)
        {
            printf("Case %d: 2 %lld\n", ++kase, (ll)m * (ll)(m - 1ll) / 2ll);
            memset(size, 0, sizeof(size));
            memset(belong, 0, sizeof(belong));
            memset(dfn, 0, sizeof(dfn));
            memset(low, 0, sizeof(low));
            memset(vis, 0, sizeof(vis));
            memset(d, 0, sizeof(d));
            memset(belong, 0, sizeof(belong));
            for(int i = 1; i <= m; ++i)
                G[i].clear();
            for(int i = 1; i <= cnt; ++i)
                g[i].clear();
            cnt = 0;
            all = 0;
            m = 0;
            top = 0;        
            continue;
        }
        for(int i = 1; i <= cnt; ++i)
        {
            sort(g[i].begin(), g[i].end());
            g[i].erase(unique(g[i].begin(), g[i].end()), g[i].end());
            for(int j = 0; j < g[i].size(); ++j)
            {
                ++d[i];
                ++d[g[i][j]];
            }            
        }
        ll ans1 = 0, ans2 = 1;
        for(int i = 1; i <= cnt; ++i)
            if(d[i] == 2)
            {
                ++ans1;
                ans2 *= size[i];
            }
        printf("Case %d: %lld %lld\n", ++kase, ans1, ans2);
        memset(size, 0, sizeof(size));
        memset(belong, 0, sizeof(belong));
        memset(dfn, 0, sizeof(dfn));
        memset(low, 0, sizeof(low));
        memset(vis, 0, sizeof(vis));
        memset(d, 0, sizeof(d));
        memset(belong, 0, sizeof(belong));
        for(int i = 1; i <= m; ++i)
            G[i].clear();
        for(int i = 1; i <= cnt; ++i)
            g[i].clear();
        cnt = 0;
        all = 0;
        m = 0;
        top = 0;
    }
//    fclose(stdin); fclose(stdout);
    return 0;
}