POJ 3895 Cycles of Lanes (dfs)

POJ 3895 Cycles of Lanes (dfs)

Description

Each of the M lanes of the Park of Polytechnic University of Bucharest connects two of the N crossroads of the park (labeled from 1 to N). There is no pair of crossroads connected by more than one lane and it is possible to pass from each crossroad to each other crossroad by a path composed of one or more lanes. A cycle of lanes is simple when passes through each of its crossroads exactly once.
The administration of the University would like to put on the lanes pictures of the winners of Regional Collegiate Programming Contest in such way that the pictures of winners from the same university to be on the lanes of a same simple cycle. That is why the administration would like to assign the longest simple cycles of lanes to most successful universities. The problem is to find the longest cycles? Fortunately, it happens that each lane of the park is participating in no more than one simple cycle (see the Figure).

Input

On the first line of the input file the number T of the test cases will be given. Each test case starts with a line with the positive integers N and M, separated by interval (4 <= N <= 4444). Each of the next M lines of the test case contains the labels of one of the pairs of crossroads connected by a lane.

Output

For each of the test cases, on a single line of the output, print the length of a maximal simple cycle.

Sample Input

1 
7 8 
3 4 
1 4 
1 3 
7 1 
2 7 
7 5 
5 6 
6 2

Sample Output

4

Source

Southeastern European Regional Programming Contest 2009


題意就是求最大的環包含點的個數，但是有一個麻煩的地方，就是dfs 到一個點時，怎麼快速知道與這個點相連點呢？ 後來我知道了 vector 開一個vector數組保存就行了，接下來就是代碼了，細節在代碼中

#include
#include
#include
#include
#include
using namespace std;
#define N 5000

vectorq[N];

int vis[N];
int n,m;
int ans;

void dfs(int a,int pos)
{
    int i;
    vis[a]=pos;

    for(i=0;ians)  //例如環 1 2 3 4 1 ,vis[1]=1,vis[4]=4,下一次4與
                                    //1相連，但是已經vis[1],所以環的大小 vis[4]-vis[1]+1
               ans=vis[a]-vis[x]+1;
     }
}

int main()
{
    int i,t;
    scanf("%d",&t);

    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(vis,0,sizeof(vis));
        int x,y;
        
        for(i=1;i<=n;i++)  //記得清空上次的數據
            q[i].clear();
        
        while(m--)
        {
            scanf("%d%d",&x,&y);
            q[x].push_back(y);  //q[x]存與x的臨邊
            q[y].push_back(x);  //同理
        }

        ans=0;
       for(i=1;i<=n;i++)
        if(!vis[i])  //因為一個點就會出現一次，即使兩個環有共同邊，
                     //那麼在公共邊那個分叉點還是會分別進行dfs的
            dfs(i,1);
                    //不加下面會錯
        if(ans<=2)//一個環需要3個點，但是我郁悶了，如果沒有環的話
                  //怎麼會有vis[x]已經標記了呢？（此時ans=0啊）難道有數據類似
                  // a b   b a  (⊙o⊙)…
            ans=0;
        printf("%d\n",ans);
    }
    return 0;
}

dfs還不是很熟，但是有事纏身，哎以後一定好好補補dfs