CUC DFS及其应用 - Virtual Judge

题目大意：有一个n个点的图，求每个点如果被删除后，图中剩余连通块的数量

1<=n<=1e4

思路：也就是求每一个割点出现在几个双联通分量（没有割点的图）中，找双连通分量也是用tarjan，dfn[u]表示u是第几个遍历到的点，low[u]表示从u出发在不经过父子边的情况下能到达的最早的祖先的dfn值，而如果一个点是割点，那么在不经过该点的情况下无法到达该点的祖先也就是low[v]>=dfn[u]，注意特判起始点的情况，它可能是假割点

//#include<__msvc_all_public_headers.hpp>
#include<bits/stdc++.h>
using namespace std;
const int N = 1e4 + 5;
int head[N], head2[N];
struct Edge
{int v, next;
}e[N*10];
int cnt = 0;
int n,m;
int dfn[N];
void addedge(int u, int v)
{e[++cnt].v = v;e[cnt].next = head[u];head[u] = cnt;
}
struct Node
{int id, cn;
}node[N];
int tot = 0;
void init()
{tot = 0;cnt = 0;for (int i = 0; i <= n; i++){node[i].id = i;node[i].cn = 1;//答案至少是1head[i] = -1;dfn[i] = 0;}
}
int tarjan(int u,int fa)
{int lowu = dfn[u] = ++tot;int child = 0;for (int i = head[u]; ~i; i = e[i].next){int v = e[i].v;if (!dfn[v]){child++;int lowv = tarjan(v, u);lowu = min(lowu, lowv);if (lowv >= dfn[u]){//该点是割点node[u].cn++;}}else if (dfn[v] < dfn[u] && v != fa){lowu = min(lowu, dfn[v]);}}if (fa < 0 && child == 1){//假割点node[u].cn = 1;}return lowu;
}
bool cmp(Node a, Node b)
{return a.cn == b.cn ? a.id<b.id : a.cn>b.cn;
}
void solve()
{init();int u, v;while(cin>>u>>v){if (u == -1 && v == -1){break;}addedge(u, v);addedge(v, u);}tarjan(0, -1);sort(node, node + n, cmp);for (int i = 0; i < m; i++){cout << node[i].id << " " << node[i].cn << endl;}cout << endl;
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);int t;//cin >> t;while (cin>>n>>m){if (!n&&!m){break;}solve();}return 0;
}