http://codeforces.com/problemset/problem/999/E

題意 有向圖 給你n個點，m條邊，以及一個初始點s，問你至少還需要增加多少條邊，使得初始點s與剩下其他的所有點都連通。

第一個想法自然是通過上標記的方法，對每一個入度為0的點跑dfs。

但是問題在於剩下沒有上標記的點，是成環的點。這些點不能有效的形成我們希望的拓撲序。

第一個想法是可以考慮上特殊標記，順序枚舉每個環上的點跑dfs，對每個隨機跑的點上標記，在dfs的過程中如果可以經過之前枚舉跑到的起點，就去掉這個點的標記，隨後統計特殊標記的數量，經過測試，確實可以AC

#include <map>
#include 
 
<set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include 
 
<functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define
 
 CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
const double eps = 1e-9;
const int maxn = 5010;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,tmp,K,s; 
bool vis[maxn];
bool vis2[maxn];
bool vis3[maxn];
int ind[maxn];
struct Edge{
    int to,next;
}edge[maxn];
int head[maxn],tot,ans;
void init(){
    Mem(head,0);
    tot = 0;
}
void add(int u,int v){
    edge[++tot].next = head[u];
    edge[tot].to = v;
    head[u] = tot;
}
void dfs(int t){
    vis[t] = 1;
    for(int i = head[t]; i;i = edge[i].next){
        int v = edge[i].to;
        if(vis[v]) continue;
        dfs(v);
    }
}
void dfs2(int t){
    vis3[t] = 1;
    vis[t] = 1;
    for(int i = head[t];i;i = edge[i].next){
        int v = edge[i].to;
        if(vis2[v]){
            vis2[v] = 0;
            ans--;
        }
        if(vis3[v]) continue;
        dfs2(v);
    }
}
int main()
{
    scanf("%d%d%d",&N,&M,&s);
    init();
    For(i,1,M){
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v);
        ind[v]++;
    }
    dfs(s);
    ans = 0;
    For(i,1,N){
        if(!ind[i] && !vis[i]){
            ans++;
            dfs(i);
        }
    }
    For(i,1,N){
        if(!vis[i]){
            Mem(vis3,0);
            ans++;
            dfs2(i);
            vis2[i] = 1;
        }
    }
    Pri(ans);
    #ifdef VSCode
    system("pause");
    #endif
    return 0;
}

第二個想法是標記覆蓋，順序dfs每一個點，以每一個點為起點經過的點用不同標記覆蓋，最後判斷所有標記的數量

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
const double eps = 1e-9;
const int maxn = 5010;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,tmp,K,s; 
int vis[maxn];
bool vis2[maxn];
bool vis3[maxn];
int ind[maxn];
struct Edge{
    int to,next;
}edge[maxn];
int head[maxn],tot,ans;
void init(){
    Mem(head,0);
    tot = 0;
}
void add(int u,int v){
    edge[++tot].next = head[u];
    edge[tot].to = v;
    head[u] = tot;
}
void dfs(int t){
    vis[t] = tmp;
    for(int i = head[t]; i;i = edge[i].next){
        int v = edge[i].to;
        if(vis[v] == tmp) continue;
        dfs(v);
    }
}
int main()
{
    scanf("%d%d%d",&N,&M,&s);
    init();
    For(i,1,M){
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v);
    }
    tmp = 1;
    dfs(s);
    For(i,1,N){
        if(!vis[i]){
            tmp++;
            dfs(i);
        }
    }
    ans = 0;
    vis2[1] = 1;
    For(i,1,N){
        if(!vis2[vis[i]]){
            vis2[vis[i]] = 1;
            ans++;
        }
    }
    Pri(ans);
    #ifdef VSCode
    system("pause");
    #endif
    return 0;
}

當然，以上兩個算法都是O(n2)的算法，這題5000的數據範圍可以跑，但是當數據範圍擴大的時候，就需要考慮更強的算法來解決；

用Tarjan算法將原圖縮點變成一個可拓撲的dag圖，直接數入度為0的點即可。

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
const double eps = 1e-9;
const int maxn = 5010;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,tmp,K,s; 
int vis[maxn];
struct Edge{
    int to,next;
}edge[maxn];
int head[maxn],tot,ans;
int Low[maxn],dfn[maxn],Stack[maxn],belong[maxn];
int index,top,scc;
bool Instack[maxn];
int ind[maxn];
int num[maxn];
void Tarjan(int u){
    int v;
    Low[u] = dfn[u] = ++ index;
    Stack[top++] = u;
    Instack[u] = true;
    for(int i = head[u];i;i =edge[i].next){
        int v = edge[i].to;
        if(!dfn[v]){
            Tarjan(v);
            if(Low[u] > Low[v]) Low[u] = Low[v];
        }else if(Instack[v] && Low[u] > dfn[v]){
            Low[u] = dfn[v];
        }
    }
    if(Low[u] == dfn[u]){
        scc++;
        do{
            v = Stack[--top];
            Instack[v] = false;
            belong[v] = scc;
            num[scc]++;
        }while(v != u);
    }
}
void init(){
    Mem(head,0);
    tot = 0;
}
void add(int u,int v){
    edge[++tot].next = head[u];
    edge[tot].to = v;
    head[u] = tot;
}

int main()
{
    scanf("%d%d%d",&N,&M,&s);
    init();
    For(i,1,M){
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v);
    }
    index = scc = top = 0;
    For(i,1,N) if(!dfn[i]) Tarjan(i);
    int ans = 0;
    For(i,1,N){
        for(int j = head[i];j;j = edge[j].next){
            int v = edge[j].to;
            if(belong[i] != belong[v]){
                ind[belong[v]] = 1;
            }
        }
    }
    vis[belong[s]] = 1;
    For(i,1,N){
        if(!ind[belong[i]] && !vis[belong[i]]){
            vis[belong[i]] = 1;
            ans++;
        }
    }
    Pri(ans);
    #ifdef VSCode
    system("pause");
    #endif
    return 0;
}

CodeForces999E 雙dfs // 標記覆蓋 // tarjan縮點