ZOJ - 3816 Generalized Palindromic Number dfs

Generalized Palindromic Number

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Time Limit: 2 Seconds                                     Memory Limit: 65536 KB                            

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A number that will be the same when it is written forwards or backwards is known as a palindromic number. For example, 1234321 is a palindromic number.

We call a number generalized palindromic number, if after merging all the consecutive same digits, the resulting number is a palindromic number. For example, 122111 is a generalized palindromic number. Because after merging, 122111 turns into 121 which is a palindromic number.

Now you are given a positive integer N, please find the largest generalized palindromic number less than N.

Input

There are multiple test cases. The first line of input contains an integer T (about 5000) indicating the number of test cases. For each test case:

There is only one integer N (1 <= N <= 10¹⁸).

Output

For each test case, output the largest generalized palindromic number less than N.

Sample Input

4
12
123
1224
1122

Sample Output

11
121
1221
1121

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Author: LIN, Xi                                         Source: The 2014 ACM-ICPC Asia Mudanjiang Regional First Round

转自：http://blog.csdn.net/u011345136/article/details/39122741

题意：求小于N的回文数，这个数的回文相同的数可以缩成一个数
思路：dfs(l, r, flag)：l代表左边侧长度，r代表右边的长度，flag代表是否处于边界，然后在搜索右边匹配左边的同时，枚举k，k代表右边连续相同的数都匹配左边的个数

 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<map>

 #define N 55
 #define M 15
 #define mod 6
 #define mod2 100000000
 #define ll long long
 #define maxi(a,b) (a)>(b)? (a) : (b)
 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 int T;
 ll n;
 char s[N];
 ll ans;
 int len;
 char leftnum[N];
 char rightnum[N];

 void ini()
 {
 //    scanf("%lld",&n);
     cin>>n;
     ans=-;
     //memset(left,0,sizeof(left));
     //memset(right,0,sizeof(right));
     sprintf(s+,"%lld",n);
     //sprintf(s+1,"%I64d",n);
     len=strlen(s+);
     for(int i=;i<=len;i++){
         s[i]-='';
     }
 }

 void out()
 {
     //printf("%lld\n",ans);
     cout<<ans<<endl;
 }

 ll getans(int l,int r)
 {
     ll re=;
     int i;
     for(i=;i<=l;i++){
         re=re*+leftnum[i];
     }
     for(i=r;i>=;i--){
         re=re*+rightnum[i];
     }
     return re;
 }

 ll dfs(int l,int r,int flag)
 {
     ll re=-;
     int i,k,m;
     if(l+r->len) return -1ll;
     if(l+r-==len){
         re=getans(l-,r);
         if(re>=n) return -1ll;
         return re;
     }

     m= (flag==) ? s[l] : ;
     for(i=m;i>=;i--)
     {
         leftnum[l]=i;
         if( (l== || leftnum[l]!=leftnum[l-]) && !(l== && i==) && (l+r!=len) )
         {
             for(k=;k+l+r<=len;k++){
                 rightnum[k+r]=i;
                 re=max(re,dfs(l+,r+k,flag && (m==i) ) );
             }
         }
         else{
             re=max(re,dfs(l+,r,flag && (m==i) ));
         }
         if(re>){
             return re;
         }
     }
     return re;
 }

 int main()
 {
     //freopen("data.in","r",stdin);
     scanf("%d",&T);
     for(int cnt=;cnt<=T;cnt++)
    // while(T--)
     //while(scanf("%I64d%I64d%I64d",&a,&b,&c)!=EOF)
     {
         ini();
         ans=dfs(,,);
         out();
     }

     return ;
 }