uva10465 (full backpack, requires full backpack)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category= 114&problem=1406&mosmsg=submission+received+with+id+15392606

Give us the cost of two items, and the capacity of a backpack

Ask us to take as many items as possible with the least amount of wasted backpack

That is, the full knapsack problem is required, as long as the attention is initialized, the dp[0] is set to 0, the other dp[i] is set to 1, indicating illegal

Then a state must be transferred from a legitimate one, and then the maximum value is selected in the legal State transfer

DP, we enumerate the capacity in reverse order to find a valid DP state

Then the DP state is a waste of the least capacity state

1#include <stdio.h>2#include <string.h>3#include <stdlib.h>4#include <algorithm>5#include <iostream>6#include <queue>7#include <stack>8#include <vector>9#include <map>Ten#include <Set> One#include <string> A#include <math.h> - using namespacestd; - #pragmaWarning (disable:4996) thetypedefLong LongLL;  - Const intINF =1<< -; - /* -  + */ - inta[2]; + intdp[10000+Ten]; A intMain () at { -     intN, I, J; -      while(SCANF ("%d%d%d", &a[0], &a[1], &n)! =EOF) -     { -          for(i =1; I <= N; ++i) -Dp[i] =-1; indp[0] =0; -          for(i =0; I <2; ++i) to          for(j = a[i]; J <= N; + +j) +         { -             if(Dp[j-a[i]]! =-1)//Select the maximum value in a legitimate state transfer theDP[J] = max (Dp[j], Dp[j-a[i]] +1); *         } $          for(j = n; J >=1; --j)Panax Notoginseng             if(Dp[j]! =-1)//find a legal state . -              Break; the         if(J = =N) +printf"%d\n", Dp[j]); A         Else theprintf"%d%d\n", Dp[j], N-j); +     } -     return 0; $}

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uva10465 (full backpack, requires full backpack)