01 knapsack problem

Reference to the Http://www.cnblogs.com/qinyg/archive/2012/04/26/2471829.html

The key state transfer equations are as follows

V (i,j) indicates that the maximum value of an item in a backpack with a capacity of J (1<=j<=c) can be loaded in the first I (1<=I<=N) item, and the following dynamic planning functions are available:

(1) V (i,0) =v (0,j) =0

(2) V (i,j) =v (i-1,j) J<wi

V (i,j) =max{v (i-1,j), V (I-1,J-WI) +vi)} J>wi

(1) type indicates: If the weight of the item I is greater than the capacity of the backpack, then the maximum value that can be obtained by the front I item is the same as the maximum price to be loaded before the i-1 item, i.e. the item I cannot be loaded into the backpack; the first (2) formula indicates that if the weight of item I is less than the capacity of the backpack, there are (a) If the item I is loaded into a backpack, the value of the backpack item is equal to the value of the J-WI in the backpack of the i-1 item, plus the value of the article I, VI; (b) If item I is not loaded into a backpack, the value of the item in the backpack is equal to the value obtained in carrying the first i-1 item into a backpack of capacity J. Obviously, taking the most value of the two is the best solution for loading the first I item into a backpack with a capacity of J.

With poj3624 practiced hand

First open two-dimensional DP array, direct hyper memory

The second time alternating with two one-dimensional arrays, memory is not super, but it takes more than 400 MS, and the code is ugly

1#include <iostream>2#include <string>3#include <sstream>4#include <vector>5 6 using namespacestd;7 8 9 intMain ()Ten { One     intN, M; ACIN >> N >>m; -     int*w = (int*)malloc(sizeof(int) * (n+1)); -     int*d = (int*)malloc(sizeof(int) * (n +1)); the  -     int*DP = (int*)(malloc(sizeof(int) * (m+1))); -     int*dp_t = (int*)(malloc(sizeof(int) * (M +1))); -  +Memset (DP,0,sizeof(int) * (M +1)); -  +      for(inti =1; I <= N; i++) A     { atCIN >> W[i] >>D[i]; -     } -  -      for(inti =1; I <= N; i++) -     { -memset (dp_t,0,sizeof(int) * (M +1)); in  -          for(intj =1; J <= M; J + +) to         { +             intx =Dp[j]; -             inty =0; the             if(J >=W[i]) *y = Dp[j-w[i]] +D[i]; $DP_T[J] = x < y?y:x;Panax Notoginseng              -         } the  +memcpy (DP, dp_t,sizeof(int) * (M +1)); A     } the  +cout << Dp[m] <<Endl; -  $     //System ("pause"); $}

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Looked at someone else's code, only to find a more elegant processing method, the revised code as follows, time-consuming is reduced to less than 300ms

1#include <iostream>2#include <string>3#include <sstream>4#include <vector>5 6 using namespacestd;7 8 intdp[12900];9 intw[3410], d[3410];Ten  One intMain () A { -     intN, M; -CIN >> N >>m; the  -Memset (DP,0,sizeof(DP)); -  -      for(inti =1; I <= N; i++) +     { -CIN >> W[i] >>D[i]; +     } A  at      for(inti =1; I <= N; i++) -     { -  -          for(intj = m; J >0; j--) -         { -             intx =Dp[j]; in             inty =0; -             if(J >=W[i]) toy = Dp[j-w[i]] +D[i]; +DP[J] = x < y?y:x; -         } the     } *  $cout << Dp[m] <<Endl;Panax Notoginseng  -     //System ("pause"); the}

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01 knapsack problem