题目传送门

LightOJ 1026

LightOJ 1063

其实是一个模板题，使用Tarjan算法来处理出桥和割点。
dfn[i]是顶点i的时间戳，low[i]是顶点i能够回到的最早的祖先。

割点

• 是树根，并且可处理的孩子数量>1
• 不是树根，但是其某个孩子能够访问的最早的祖先的时间戳大于或者等于该点的时间戳

割边

• 有一顶点u，它的某个孩子v能够访问的最早的祖先的时间戳大于u的时间戳，那么由u和v所组成的边是割边。

代码：

Light OJ 1026

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#include <string>
#include <math.h>
using namespace std;
typedef pair<int,int> P;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-9;
const int N = 1e4 + 5;
map<int,int> mp;
struct Edge
{int u,v;int cut;Edge(){}Edge(int _u, int _v):u(_u), v(_v) {cut = 0;}
};
vector<Edge> edges;
vector<int> G[N];
int bridges;
int dfn[N],low[N],dfs_clock;void init()
{for(int i = 0; i < N; i++)G[i].clear();edges.clear();mp.clear();memset(dfn, 0, sizeof(dfn));memset(low, 0, sizeof(low));bridges = 0;dfs_clock = 0;
}bool Hash(int u, int v)
{int tmp = u*10000 + v;if(mp.count(tmp)) return true;mp[tmp] = 1;mp[v*10000+u] = 1;return false;
}void addedge(int u, int v)
{edges.push_back(Edge(u,v));edges.push_back(Edge(v,u));int m = edges.size();G[u].push_back(m-2);G[v].push_back(m-1);
}void dfs(int u, int fa)
{dfn[u] = low[u] = ++dfs_clock;for(int i = 0; i < G[u].size(); i++){Edge &e = edges[G[u][i]];int v = e.v;if(v == fa) continue;if(!dfn[v]){dfs(v,u);low[u] = min(low[u], low[v]);}else{low[u] = min(low[u], dfn[v]);}if(low[v] > dfn[u]){bridges++;e.cut = 1;edges[G[u][i]^1].cut = 1;}}
}int t,kase = 0;
int u,v,m,n;
int main()
{scanf("%d", &t);while(t--){scanf("%d", &n);init();for(int i = 0; i < n; i++){scanf("%d (%d)", &u, &m);for(int j = 0; j < m; j++){scanf("%d", &v);if(!Hash(u,v))addedge(u,v);}}for(int i = 0; i < n; i++)if(!dfn[i])dfs(i,-1);printf("Case %d:\n", ++kase);printf("%d critical links\n", bridges);vector<P> ans;for(int i = 0; i < edges.size(); i++){if(edges[i].cut && edges[i].u < edges[i].v){ans.push_back(make_pair(edges[i].u,edges[i].v));}}sort(ans.begin(),ans.end());for(int i = 0; i < ans.size(); i++){printf("%d - %d\n", ans[i].first, ans[i].second);}}return 0;
}

Light OJ 1063

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#include <string>
#include <math.h>
using namespace std;
typedef pair<int,int> P;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-9;
const int N = 1e4 + 5;
struct Edge
{int u,v;Edge(){}Edge(int a, int b):u(a),v(b) {}
};
vector<Edge> edges;
vector<int> G[N];
int cut[N];
int dfn[N], low[N], dfs_clock;
int n,m;
int ans = 0;
void init()
{edges.clear();for(int i = 0; i < N; i++) G[i].clear();memset(cut, 0, sizeof(cut));memset(dfn, 0, sizeof(dfn));memset(low, 0, sizeof(low));dfs_clock = 0;
}void addedge(int u, int v)
{edges.push_back(Edge(u, v));edges.push_back(Edge(v, u));int m = edges.size();G[u].push_back(m-2);G[v].push_back(m-1);
}void dfs(int u, int fa)
{dfn[u] = low[u] = ++dfs_clock;int son = 0;for(int i = 0; i < G[u].size(); i++){Edge &e = edges[G[u][i]];int v = e.v;if(v == fa) continue;if(!dfn[v]){son++;dfs(v, u);low[u] = min(low[u], low[v]);if(u != fa && low[v] >= dfn[u]){cut[u] = 1;}}elselow[u] = min(low[u], dfn[v]);}if(u == fa && son > 1){cut[u] = 1;}}
int t,kase=0;
int main()
{scanf("%d", &t);while(t--){init();scanf("%d%d", &n,&m);for(int i = 0; i < m; i++){int u,v;scanf("%d%d", &u,&v);addedge(u,v);}ans = 0;for(int i = 1; i <= n; i++)if(!dfn[i]) dfs(i, i);for(int i = 1; i <= n; i++)if(cut[i]) ans++;printf("Case %d: %d\n", ++kase, ans);}return 0;
}