Hdoj 2191 mourn 512 .. [Multiple backpacks] + [binary decomposition]

The question is really long...

Question: Chinese question, you know ..

Policy: Multiple knapsack problems.The problem of converting multiple backpacks to 01 is that multiple backpacks are initialized, and the number of its parts is a power of 2.The number can be combined into a set of several pieces, which can be any number smaller than or equal to C and will not be repeated. It is called binary decomposition because it can be decomposed in this way.

It is interpreted in the binary form of a number.

For example, if the binary value of 7 is 7 = 111, it can be divided into 001, 010, 100. Combine them into any number less than or equal to 7, and each combination gets a different number. 15 = 1111 can be divided into 0001 0010 0100 1000 four numbers If 13 = 1101, it is divided into 0001 0010 0100. The first three digits can be combined Any number within 7, plus 0110 = 6 can be combined into any one greater than 6 less than 13 Although there are duplicates, we can always consider all the numbers within 13. Thought (Binary thinking) to convert multiple items into a variety of items, you can use 01 backpack to solve the problem.Link: Click to open the link code:

# Include <stdio. h> # include <string. h> # include <math. h> int DP [555], W [555], V [555]; // W indicates the quality, and V indicates the money to be spent. Int main () {int t, n, m, P, h, C, cou; scanf ("% d", & T); While (t --) {scanf ("% d", & N, & M); memset (DP, 0, sizeof (DP); cou = 0; int I, j; while (M --) {scanf ("% d", & P, & H, & C);/* for (I = 1; I <= C; I + = (INT) Pow (2.0, I) {// POW is always wa. Check that POW returns not int type, but double type, although the test data can pass, however, it may be wrong to get an integer. Therefore, we use an integer to determine V [cou] = I * P; W [cou ++] = I * h; C-= I ;} */int l = 1; while (C> = L) {v [cou] = L * P; W [cou ++] = L * h; C-= L; L <= 1 ;}if (c) {v [cou] = C * P; W [cou ++] = C * H ;}} for (I = 0; I <cou; I ++) {for (j = N; j> = V [I]; j --) {If (DP [J] <DP [J-V [I] + W [I]) DP [J] = DP [J-V [I] + W [I] ;}} printf ("% d \ n", DP [N]);} return 0 ;}