[HDU] 1269 迷宫城堡最简单的强连通分支题

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1269

方法：就是求强连通分支个数，为1就可以。

感想：《算法导论》上额强连通分支部分的定理证明有必要多理解。

代码：

#include<iostream>
#include<algorithm>
using namespace std;
//int const MAX =0x3f3f3f3f;
int const MAX = 10005;
struct Arc
{
    int vetex;
    Arc* nexttArc;
};

struct Node
{
    int x;
    Arc* firstArc;
};
Node* o_nodes[MAX];
Node* r_nodes[MAX];
bool visited[MAX];
int n,m;
void createArc(int x,int y,bool inO)
{
    Arc* arc = (Arc*)(malloc(sizeof(Arc)));
    arc->vetex = y;
    if(inO)
    {
        if(o_nodes[x]->firstArc == NULL)
            arc->nexttArc=NULL;
        else
            arc->nexttArc=o_nodes[x]->firstArc;
        o_nodes[x]->firstArc = arc;
    }
    else
    {
        if(r_nodes[x]->firstArc == NULL)
            arc->nexttArc=NULL;
        else
            arc->nexttArc=r_nodes[x]->firstArc;
        r_nodes[x]->firstArc = arc;
    }
}
void loadArc(int x,int y)
{
    createArc(x,y,true);
    createArc(y,x,false);
}
int last = 0,traversedCount=0;
void DFS(int root,bool inO=true)
{
    Arc* t_arc;
    if(inO)
        t_arc = o_nodes[root]->firstArc;
    else
        t_arc = r_nodes[root]->firstArc;
    if(!visited[root] && !inO)
        traversedCount++;

    visited[root] = true;
    while(t_arc!=NULL)
    {
        if(!visited[t_arc->vetex])
            DFS(t_arc->vetex,inO);
        t_arc = t_arc->nexttArc;
    }
    last = root; 
}
int main()
{
    while(scanf("%d %d",&n,&m) && !(n==0 && m==0))
    {  
        memset(visited,false,sizeof(visited));
        last = 0;
        traversedCount = 0;
        for(int i=1;i<=n;i++)
        {
            o_nodes[i] = (Node*)malloc(sizeof(Node));
            r_nodes[i] = (Node*)malloc(sizeof(Node));
            o_nodes[i]->firstArc = NULL;
            r_nodes[i]->firstArc = NULL;
        }
        int i = 1;
        int a,b;
        while(i<=m)
        {
            scanf("%d %d",&a,&b);
            loadArc(a,b);
            i++;
        }
        int count = 0;
        bool valid = true;
        for(i=1;i<=n;i++)
        {
            if(visited[i] == false)
            {
                if(count>0)
                {
                    valid = false;
                    break;
                }
                else
                {
                    DFS(i,true);
                    count++;
                }
            }
        }
        if(!valid)
            cout<<"No"<<endl;
        else
        {
            memset(visited,false,sizeof(visited));
            DFS(last,false);
            if(traversedCount==n)
                cout<<"Yes"<<endl;
            else
                cout<<"No"<<endl;
        }
    }
    return 0;
}