Proving Equivalences

2
4 0
3 2
1 2
1 3

4
2

题意：n个点m条边，问最少加入多少条边使得整个图联通。

思路：先Tarjan求强联通分量，缩点，再求缩点后的点的入度和出度，入读为0的点的个数为a。出度为0的点的个数为b，ans=max（a。b）

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

const int MAXN = 20050;//点数
const int MAXM = 500050;//边数

struct Edge
{
    int to,next;
}edge[MAXM];

int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong数组的值是1~scc
int Index,top;
int scc;//强联通分量的个数
bool Instack[MAXN];
int num[MAXN];//各个强联通分量包括的点的个数。数组编号为1~scc
//num数组不一定须要，结合实际情况

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

void Tarjan(int u)
{
    int v;
    Low[u]=DFN[u]=++Index;
    Stack[top++]=u;
    Instack[u]=true;
    for (int i=head[u];i+1;i=edge[i].next)
    {
        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 solve(int N)
{
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    memset(num,0,sizeof(num));
    Index=scc=top=0;
    for (int i=1;i<=N;i++)      //点的编号从1開始
        if (!DFN[i])
            Tarjan(i);
}

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

int n,m;
int in[MAXN],out[MAXN];

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
#endif
    int i,j,u,v,t;
    sf(t);
    while (t--)
    {
        sff(n,m);
        if(n==1){   //特判1（n==1,m==0)
            printf("0\n");
            continue;
        }
        if(m==0){   //特判2( n==?
,m==0)
            printf("%d\n",n);
            continue;
        }
        init();
        for (i=0;i<m;i++)
        {
            sff(u,v);
            addedge(u,v);
        }
        solve(n);
        if(scc==1){   //假设强连通个数为1
            printf("0\n");
            continue;
        }
        mem(in,0);
        mem(out,0);
        for (int u=1;u<=n;u++)
        {
            for (i=head[u];i+1;i=edge[i].next)
            {
                int v=edge[i].to;
                if (Belong[u]!=Belong[v])
                {
                    out[Belong[u]]++;
                    in[Belong[v]]++;
                }
            }
        }
        int ans,a=0,b=0;
        for (i=1;i<=scc;i++)
        {
            if (out[i]==0)
                a++;
            if (in[i]==0)
                b++;
        }
        ans=max(a,b);
        pf("%d\n",ans);
    }
    return 0;
}