POJ1014 Problem Solving report (DFS)

Title here: http://poj.org/problem?id=1014

To see test instructions, the title of the input is the value of the number of 1,2,3,4,5,6 marble, instead of 6 value for the input number of marble! Choose this topic mainly want to practice a deep search. On the internet to see a piece of code, deep search using the greedy algorithm although it can be AC but there are a lot of loopholes:

For example, test data (0 0 3 0 3 1) An error occurs:

Search time will be selected: The value of each of the 6,5,3 to determine that can not be divided. But the right way to split it should be 6+3+3+3=15.

This is a vulnerable code on the Internet:

1#include <iostream>2 using namespacestd; 3   4 intamount[7] = {0}; 5 intHalf_value =0; 6 intFlag =0; 7   8 voidDFS (intValueintpre) {  9   Ten     if(Value = =half_value) {   OneFlag =1;  A         return;  -     }   -    the     if(Flag = =1){//This can be removed from the AH!  -         return;  -     }   -    +     inti =0;  -      for(i = pre; I >0; i--){   +         if(Amount[i]) { A             if(i + Value <=half_value) {   atamount[i]--;  -DFS (i +value, i);  -    -                 if(Flag = =1){//no less, feel its role, let the recursive stack end all DFS -                     return;  -                 }   in             }   -         }   to     }   + }   -    the    * intMain () { $   Panax Notoginseng     intTestCase =1;  -      while(true){   theFlag =0;  +         intTotalvalue =0;  A         intN =6;  the         inti =1;  +          while(I <=N) {   -CIN >>Amount[i];  $Totalvalue + = amount[i] *i;  $i++;  -         }   -    the         if(!amount[1] &&!amount[2] &&!amount[3] &&!amount[4] &&!amount[5] &&!amount[6]){   -              Break; Wuyi         }   the    -printf"Collection #%d:\n", testcase++);  Wu         if(Totalvalue%2!=0){   -cout <<"Can ' t be divided."<< Endl <<Endl;  About             Continue;  $         }   -    -Half_value = Totalvalue/2;  -Daso0,6);  A    +         if(flag) { thecout <<"Can be divided."<<Endl;  -}Else {   $cout <<"Can ' t be divided."<<Endl;  the         }   thecout <<Endl;  the     }      the     return 0;  -}

a vulnerable Code

In order to solve this problem, before the decision can not be divided into a cycle, that is, from the value of 5,4...1 marble began deep search, so as to avoid the above code loopholes. Because the greedy algorithm basically solves most of the situation, the loop does not cause a timeout. The following is the modified code:

#include <iostream>#include<string.h>#include<stdio.h>using namespacestd;intamount[7] = {0 };inttemp_amount[7] = {0};//Save the original input valueintHalf_value =0;intFlag =0;voidDasointValueintpre) {    if(Value = =half_value) {Flag=1; return; }    inti =0;  for(i = pre; I >0; i--)    {        if(Amount[i]) {if(i + Value <=half_value) {Amount[i]--; DFS (i+value, i); if(Flag = =1)                {                      return; }            }        }    }}intMain () {intTestCase =1;  while(true) {flag=0; intTotalvalue =0; intN =6; inti =1;  while(I <=N) {cin>>Temp_amount[i]; Totalvalue+ = temp_amount[i] *i; I++; }        if(!temp_amount[1] &&!temp_amount[2] &&!temp_amount[3] &&!temp_amount[4] &&!temp_amount[5] &&!temp_amount[6])        {             Break; } printf ("Collection #%d:\n", testcase++); if(Totalvalue%2!=0) {cout<<"Can ' t be divided."<< Endl <<Endl; Continue; } half_value= Totalvalue/2; memcpy (Amount, Temp_amount,7*sizeof(int)); DFS (0,6); if(flag) cout<<"Can be divided."<<Endl; Else        {             for(intm =6; M >=1; m--)            {                if(!flag) {memcpy (amount, Temp_amount,7*sizeof(int));//because the amount array has already been manipulated, it needs to be restoredDaso0, M); }            }            if(!flag) cout<<"Can ' t be divided."<<Endl; Elsecout<<"Can be divided."<<Endl; } cout<<Endl; }    return 0;}

(To be supplemented later, uh)

POJ1014 Problem Solving report (DFS)