poj 1659 Frogs' Neighborhood

poj 1659 Frogs' Neighborhood

http://acm.pku.edu.cn/JudgeOnline/problem?id=1659
Frogs' Neighborhood

这个题我做的很郁闷方法是对的，但是过样例一直有问题
行久才找出问题，就是在回溯的时候，还原变量的值时没
多想，弄错了，哎，都怪自己太马虎，调了好长时间！

Algorithm: 贪心
怎么谈呢？
具体做法是，每次把点按边的多少排序，然后每次选一个边
最多的点P1，让这个点和其他点P2相连，选取其他点P2的顺序也是
按顺序选，也就是按边的多少。然后一直下去，若最后构造
不出合法的图，就回溯(注意还原变量的值)，调整第二个点P2的选取.

当然有一种情况在读入数据后可以直接判断，就是当所有的边数
之后为奇数时，不能构成图.
代码：

Source Code

Problem: 1659		User: lovecanon
Memory: 248K		Time: 0MS
Language: C++		Result: Accepted

      #include < iostream >
       using   namespace  std;
       struct  node{
           int  id;
           int  edge;
      }Edge[ 11 ];
       int  map[ 11 ][ 11 ],n;
       int  cmp( const   void   * a, const   void   * b){
           struct  node  * s = (node  * )a;
           struct  node  * t = (node  * )b;
           return  t -> edge - s -> edge;
      }
       int  solve(){
          qsort(Edge + 1 ,n, sizeof (Edge[ 0 ]),cmp);
           if (Edge[ 1 ].edge == 0 )  return   1 ;
           else {
               int  i,j,u,v;
               for (i = 2 ;i <= n;i ++ )
               if (map[Edge[ 1 ].id][Edge[i].id] == 0 && Edge[i].edge != 0 ){
                  
                  map[Edge[ 1 ].id][Edge[i].id] = 1 ;map[Edge[i].id][Edge[ 1 ].id] = 1 ;
                  Edge[ 1 ].edge -- ;Edge[i].edge -- ;
                  u = Edge[ 1 ].id;v = Edge[i].id;
                   if (solve())  return   1 ;
                   else {
                       for (j = 1 ;j <= n;j ++ ) {
                           if (Edge[j].id == u) Edge[j].edge ++ ;
                           else   if (Edge[j].id == v) Edge[j].edge ++ ;
                      }
                      map[u][v] = 0 ;map[v][u] = 0 ;
                      qsort(Edge + 1 ,n, sizeof (Edge[ 0 ]),cmp);
                  }
              } 
               return   0 ;
          }
      }
       int  main(){
           int  t;
          scanf( " %d " , & t);
           while (t -- ){
               int  i,j,sum = 0 ;
              scanf( " %d " , & n);
               for (i = 1 ;i <= n;i ++ ) {
                  scanf( " %d " , & Edge[i].edge);
                  Edge[i].id = i;
                  sum += Edge[i].edge;
              }
               if (sum % 2 ) {printf( " NO\n\n " ); continue ;}
              memset(map, 0 , sizeof (map));
               if ( ! solve()) {printf( " NO\n\n " ); continue ;}  
              cout << " YES " << endl;
               for (i = 1 ;i <= n;i ++ ){
                   for (j = 1 ;j <= n;j ++ ){
                       if (map[i][j] == 1 ) printf( " 1  " );
                       else  printf( " 0  " );
                  }
                  cout << endl;
              }
              cout << endl;
          }
           return   0 ;
      }