HDU1258 DFS

题意：求n个数中的某些数的和等于t，并输出
dfs
记录下已经输出过的，然后每次比较一下，这样就能避免重复
View Code
  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<stdlib.h>
  4 #include<algorithm>
  5 using namespace std;
  6 const int maxn = 15;
  7 const int maxm = 500005;
  8 int num[ maxn ];
  9 int vis[ maxn ];
 10 int ans[ maxn ];
 11 int t,n;
 12 int flag;
 13 int new_start;
 14 struct node{
 15     int cnt;
 16     int ans[ maxn ];
 17 }my[ maxm ],temp;
 18 int cnt_my;
 19 int judge( ){
 20     sort( temp.ans,temp.ans+temp.cnt );
 21     for( int i=0;i<cnt_my;i++ ){
 22         if( temp.cnt==my[i].cnt ){
 23             int k;
 24             int fff=1;
 25             for( k=0;k<temp.cnt;k++ ){
 26                 if( temp.ans[k]==my[i].ans[k] )
 27                 {}
 28                 else{
 29                     fff=-1;
 30                     break;
 31                 }
 32             }
 33             if( fff==1 )
 34                 return i;//exist
 35         }
 36     }
 37     for( int i=0;i<temp.cnt;i++ ){
 38         my[ cnt_my ].ans[ i ]=temp.ans[i];
 39     }
 40     my[ cnt_my ].cnt=temp.cnt;
 41     cnt_my++;
 42     return -1;
 43 }
 44 int judge2(int p){
 45     int sum=0;
 46     for( int i=p;i<=n;i++ ){
 47         sum+=num[i];
 48     }
 49     return sum;
 50 }
 51 void dfs( int pos,int sum,int cnt ){
 52     if( sum==t ){
 53         temp.cnt=cnt;
 54         for( int i=0;i<cnt;i++ ){
 55             temp.ans[i]=ans[i];
 56         }
 57         if( judge()==-1 ){
 58             flag=1;
 59             for( int i=0;i<cnt;i++ ){
 60                 if( i==0 ) printf("%d",ans[i]);
 61                 else printf("+%d",ans[i]);
 62             }
 63             printf("\n");
 64         }
 65         //new_start++;
 66         return ;
 67     }
 68     for( int i=new_start;i<=n;i++ ){
 69         if( vis[ i ]==0&&(sum+num[i])<=t ){
 70             int ttt=judge2( i );
 71             for( int k=0;k<cnt;k++ )
 72                 ttt+=ans[k];
 73             if( ttt<t ) return ;
 74             vis[ i ]=1;
 75             ans[ cnt ]=num[i];
 76             dfs( i,sum+num[i],cnt+1 );
 77             vis[ i ]=0;
 78         }
 79     }
 80     return ;
 81 }
 82 
 83 int main(){
 84     while( scanf("%d%d",&t,&n),n ){
 85         for( int i=1;i<=n;i++ ){
 86             scanf("%d",&num[ i ]);
 87             vis[i]=0;
 88         }
 89         flag=-1;
 90         cnt_my=0;
 91         printf("Sums of %d:\n",t);
 92         for( int i=1;i<=n;i++ ){
 93             if( num[i]>t ) {
 94                 vis[i]=1;
 95                 continue;
 96             }
 97             if( num[i]==t&&flag==-1 ){//在这个判断条件上wa了一次！！！flag！=-1
 98                 flag=1;
 99                 vis[i]=1;
100                 my[ 0 ].cnt=1;
101                 my[ 0 ].ans[0]=t;
102                 cnt_my++;
103                 printf("%d\n",t);
104                 continue;
105             }
106             if( num[i]==t&&flag==1 ){
107                 vis[i]=1;
108                 continue;
109             }
110             new_start=i;
111             dfs( i,0,0 );//pos sum cnt
112             break;
113         }
114         if( flag==-1 )
115             printf("NONE\n");
116         //printf("end\n");
117     }
118     return 0;
119 }
keep moving...