"Algorithm Design and Analysis" 7, 0/1 knapsack problem, dynamic planning

/*** Book: "Algorithmic Analysis and Design" * Function: given n items and a backpack, the weight of item I is WI, its value is VI, ask how to choose the items loaded into the backpack, so that the total value of the items loaded into the backpack? * File: beibao.cpp* time: November 30, 2014 19:22:47* cutter_point*/#include <iostream> #define Sizebeibao 20using namespace std;//The optimal substructure of this knapsack problem is/* First here there are altogether m kinds of goods, the total capacity of the backpack is W1, starting from the nth start to the front, the initialization of Nth is C[n][j] J is the capacity of the backpack, starting from 0 increments to the maximum capacity of the backpack W when J >= W When the value of the nth Plus to add the vn when 0 <= J < w so low n not add up that is N2, when considering the first I < N, then divided into max{c[i+1][j], C[i+1][j-wi]+vi} When J >= WI is m[i+1][j]*///we use M (i,j) to indicate the maximum value of the backpack that has been judged good 1:i-1 sequence, and the remaining capacity of the backpack is J, the item I is judged//n is the number, W is the backpack maximum capacity, c[n, W] It's a backpack. Under n Items the backpack size is the optimal value of W, V is the value of each, WI is the weight of each int beibao (int n, int w, int c[sizebeibao][sizebeibao], int *v, int *wi) {c[ Sizebeibao][sizebeibao] = {0};//Initialize all values//start with the nth start calculation, initialize nth for (int i = 0; I <= w; ++i)//Capacity weight 0 to W initialization {if (I >= wi[ N]//and the nth to compare that weight, as long as the weight exceeds the value is vn{c[n][i] = v[n];} Else{c[n][i] = 0;//backpack capacity is not reached, then the best value can only be 0}}//here from the nth//goods from the nth back iteration/*2, when considering the first I < N, then divided into max{c[i+1][j], C[i+1][j-wi ]+vi} When J >= Wi C[i+1][j-wi]+vi otherwise is m[i+1][j]*/for (int i = n-1; I >= 0; i) {for (int j = 0; J <= W; ++j)//backpack capacity increased from small to maximum {if (J >= wi[i])//When weight reaches this requirement c[i][j] = C[i + 1][j-wi[i]] + v[i];// The value of the item I get when I add a backpack elsec[i][j] = c[i + 1][j];//When I do not add a backpack}}//the final optimal solution is return c[1][w];/*//when the number of items is from 1 to n for (int i = 1; I &l t;= N; ++i) {C[i][0] = 0;//The value of the backpack corresponding to the 0 capacity is definitely 0//different weights will have different knapsack optimal value for (int wx = 1; wx <= W; ++wx) {//when the corresponding weight is increased// Here to determine whether the weight is larger than the backpack can be loaded, here the number of items I when the IF (Wi[i] > WX) c[i][wx] = C[I-1][WX];//2, when wi > W, the total amount of the first I > Total weight of c[n, w]=c[n -1, W]else//3, when n > 0 and Wi <= w c[n, w]=max{c[n-1, W], c[n-1, W-wi]+vi}{//Compare the two largest int temp = C[i-1][wx-wi [i]] + v[i];if (C[I-1][WX] > Temp) c[i][wx] = c[i-1][wx];elsec[i][wx] = temp;}}} */}//get knapsack problem vector solution void Qiux (int c[sizebeibao][sizebeibao], int *x, int *wi, int w, int n) {for (int i = 1; i < n; ++i)//from a Items to n items {if (c[i][w] = = C[i + 1][w]) x[i] = 0;//If the next item is added but the optimal value does not change, the preceding one is not added to the inside else{x[i] = 1;//If added an item value changed, So that means the new item can be added to the backpack to take up the quality w-= Wi[i];}} Whether the last item is loaded into the last remaining W to see C[N][W]Because the last c[i][w] = = C[i + 1][w] cannot judge x[n] = (c[n][w])? 1:0;} int main () {int wi[4] = {0, 4, 3, 2};int v[4] = {0, 5, 2, 1};int n = 3, W = 6;int C[sizebeibao][sizebeibao] = {0};int x [Sizebeibao] = {0};cout << "The best value is:" << endl;cout << Beibao (N, W, C, V, WI); cout << "vector x is:" &  lt;< Endl;qiux (c, X, WI, W, N); for (int i = 1; I <= n; ++i) cout << x[i] << "; cout << Endl;return 0;}

"Algorithm Design and Analysis" 7, 0/1 knapsack problem, dynamic planning