13年省赛-B题-连通分量

题意：求从1到N是否存在一条路，可以遍历每个节点。

思路：求任意两点之间是否通畅即可；

疑惑：完全暴力，bfs但是TLE，问题在于求连通分量（PS：不会）贴别人代码，先保存着。

 #include <bits/stdc++.h>
 #define CL(arr, val)    memset(arr, val, sizeof(arr))
 #define M 10007
 #define N 2007
 using namespace std;
 const int inf = 0x7f7f7f7f;
 const int mod = ;
 struct node
 {
     int v;
     int next;
 } g[M];
 int head[N],ct;
 bool ok[N][N];
 bool vt[N];
 int n,m;

 void add(int u,int v)
 {
     g[ct].v = v;
     g[ct].next = head[u];
     head[u] = ct++;
 }
 void bfs(int s)
 {
     int i;
     for (i = ; i <= n; ++i) vt[i] = false;
     vt[s] = true;
     queue<int> Q;
     Q.push(s);
     while (!Q.empty())
     {
         int u = Q.front();
         Q.pop();
         for (i = head[u]; i != -; i = g[i].next)
         {
             int v = g[i].v;
             if (!vt[v])
             {
                 vt[v] = true;
                 ok[s][v] = true;
                 Q.push(v);
             }
         }
     }
 }
 int main()
 {
 //    Read();
     int T;
     int i,j;
     scanf("%d",&T);
     int cas = ;
     while (T--)
     {
         scanf("%d%d",&n,&m);
         CL(head,-);
         ct = ;
         int x,y;
         for (i = ; i < m; ++i)
         {
             scanf("%d%d",&x,&y);
             add(x,y);
         }
         CL(ok,false);
         for (i = ; i <= n; ++i) ok[i][i] = true;
         for (i = ; i <= n; ++i) bfs(i);
         bool flag = false;
         for (i = ; i <= n && !flag; ++i)
         {
             for (j = ; j <= n && !flag; ++j)
             {
                 if (ok[i][j] || ok[j][i]) continue;
                 else
                 {
                     flag = true;
                     break;
                 }
             }
         }
         if (!flag) printf("Case %d: Kalimdor is just ahead\n",cas++);
         else printf("Case %d: The Burning Shadow consume us all\n",cas++);
     }
     return ;
 }