Optimization of space for 0-1 knapsack problems that cainiao can understand

If you do not know what the 0-1 backpack problem is, the following is a brief description of the 0-1 backpack problem. Assume that there are n items, each of which has a size of w1, w2 ...... The value of wn is v1, v2 ....... Vn. 01 A backpack is a backpack with n items taken out and several items put in total_weight space, so that the total size of the backpack is the largest and there is no optimized version for the 0-1 backpack problem, the core code above is the following [cpp] for (int I = 1; I <= n; I ++) {for (int j = 1; j <= total_weight; j ++) {if (w [I]> j) {c [I] [j] = c [I-1] [j];} else {if (c [I-1] [j]> v [I] + c [I-1] [j-w [I]) {c [I] [j] = c [I-1] [j];} else {c [I] [j] = v [I] + c [I-1] [j-w [I] ;}}} note the state transition equation c [I] [j] = max {c [I-1] [j], c [I-1] [j-w [I] + v [I]} each c [I] [j] change value only with c [I-1] [x] {x: 1... J} The c [I-1] [x] is the value saved in the previous I loop, so, c can be reduced to one-dimensional array state transition equation to c [j] = max (c [j], c [j-w [I] + v [I]); furthermore, we note that the state transition equation, each derivation of c [I] [j] is derived through c [I-1] [j-w [I, instead of using c [I] [j-w [I], the scanning sequence of j should be changed from large to small. Otherwise, the I-th time is used to evaluate the c array, the value of c [j-w [I] (that is, c [I] [j-w [I]), then calculate the value of c [j] (that is, c [I] [j]) Because j increments, then the state equation is like c [I] [j] = max (c [I-1] [j], c [I] [j-w [I] + v [I]) obviously does not match the meaning of the question. Therefore, the above Code is changed to [cpp] for (int I = 1; I <= n; I ++) {for (int j = total_weight; j> = 1; j --) {if (w [I ]> J) {c [j] = c [j]; // indicates that the I th is equal to the I-1, here, because c [j] originally saved the previous value, this does not need to be changed} else {// meaning that the weight of the I-th item is smaller than that of the backpack, therefore, you can choose whether to place item I or not if (c [j]> v [I] + c [j-w [I]) {c [j] = c [j];} else {c [j] = v [I] + c [j-w [I] ;}} finally, we can optimize it by removing unnecessary statements to complete [cpp] for (int I = 1; I <= n; I ++) optimization) {for (int j = total_weight; j> = w [I]; j --) {if (c [j] <= v [I] + c [j-w [I]) c [j] = v [I] + c [j-w [I] ;}} this beautiful code is simply unimaginable! Note that the space optimization version cannot solve the optimal sequence, but can find the optimal solution, that is, the maximum value.