SRM 437 TheSwap ( Permutation + Memory Search )

My idea:

 

    At first I solve this problem with the greedy approach (keeping the number maximal for every iteration) and all the test cases given is right . Later on I find some cases from the discussion that my code gives the wrong answer .

    For example : 35766 3

    the correct result is that 35766->75366->76356->76653

    but the greedy approach works like this : 35766->75366->76365->76635

    Here I will use the make permutation and memory search to solve this problem , and I have learn much about STL in C++ this time .

 

Problem Statement

    

There is nothing more beautiful than just an integer number.

You are given an integer n. Write down n in decimal notation with no leading zeroes, and let M be the number of written digits. Perform the following operation exactly k times:

• Choose two different 1-based positions, i and j, such that 1 <= i < j <= M. Swap the digits at positions i and j. This swap must not cause the resulting number to have a leading zero, i.e., if the digit at position j is zero, then i must be strictly greater than 1.

 

Return the maximal possible number you can get at the end of this procedure. If it's not possible to perform k operations, return -1 instead.

 

Definition

    Class:TheSwapMethod:findMaxParameters:int, intReturns:intMethod signature:int findMax(int n, int k)(be sure your method is public)      

Constraints

-n will be between 1 and 1,000,000, inclusive.-

k will be between 1 and 10, inclusive.

 

#include <vector>#include <map>#include <cmath>#include <algorithm>#include <iostream>using namespace std;//like the code make permutation and use Memory Search to solve this problemstruct TheSwap {map<pair<int,int>,int> mapp; int max(int a,int b){return a>b?a:b;}int recur(vector<int> v,int k){ //get the value of nint n=0;for (int ti=0;ti<v.size();ti++){n=n*10+v[ti];}if(k==0){return n;}if(mapp.count(make_pair(n,k)))return mapp[make_pair(n,k)];int a=-1;for(int i=0;i<v.size();i++)for(int j=i+1;j<v.size();j++){if(i==0 && v[j]==0) continue; //the first digit cannot be zeroswap(v[i],v[j]);a=max(a,recur(v,k-1));swap(v[i],v[j]);}return mapp[make_pair(n,k)] = a; //Memory Search to avoid duplicate work}int findMax(int n,int k){//change the number to vector//n=123 then v[0]=1;v[1]=2;v[2]=3;vector<int> num;while(n){num.push_back(n%10);n/=10;}for(int i=0,j=num.size()-1;i<j;i++,j--)swap(num[i],num[j]);return recur(num,k);}};