Hdoj 3535 areyoubusy (each group contains at least one backpack)

Link: (-_-) ZZ

There are N types of work sets. Each type of set has one type. 0 is to select at least one job in the Set, and 1 is to select one job in the set at most, 2. It takes a certain amount of time and happiness to make every job.

Idea: If the type is 0, it is the deformation of the group backpack. If it is 1, it is the group backpack. If it is 2, it is the 0-1 backpack. You can find it based on the type of the backpack.

PS: unlimited wa

Code:

# Include <stdio. h> # include <string. h> struct node {int T, H;} pack [5] [100002]; int C = 0, n = 0, T = 0, DP [102] [102], num [5] [102], index1 [5], IND [102]; void zeropack (INT in, int start) // select at least one {int I = 0, j = 0, K = 0, tpval = 0; for (I = 0; I <num [0] [in]; I ++) {for (j = T; j> = pack [0] [start + I]. t; j --) {If (DP [C] [J-pack [0] [start + I]. t]! =-1) DP [C] [J] = DP [C] [J]> DP [C] [J-pack [0] [start + I]. t] + pack [0] [start + I]. h? DP [C] [J]: DP [C] [J-pack [0] [start + I]. t] + pack [0] [start + I]. h; If (DP [C-1] [J-pack [0] [start + I]. t]! =-1) DP [C] [J] = DP [C] [J]> DP [C-1] [J-pack [0] [start + I]. t] + pack [0] [start + I]. h? DP [C] [J]: DP [C-1] [J-pack [0] [start + I]. t] + pack [0] [start + I]. h ;}} void onepack (INT in, int start) // You can select at most one grouping backpack. {int I = 0, j = 0, K = 0, tpval = 0; for (I = 0; I <= T; I ++) DP [C] [I] = DP [C-1] [I]; for (I = T; i> = 0; I --) {for (j = 0; j <num [1] [in]; j ++) if (pack [1] [start + J]. T <= I & DP [C-1] [I-pack [1] [start + J]. t]! =-1) DP [C] [I] = DP [C] [I]> DP [C-1] [I-pack [1] [start + J]. t] + pack [1] [start + J]. h? DP [C] [I]: DP [C-1] [I-pack [1] [start + J]. t] + pack [1] [start + J]. h ;}} void twopack (INT in, int start) // 0-1 backpack {int I = 0, j = 0, K = 0, tpval = 0; for (I = 0; I <= T; I ++) DP [C] [I] = DP [C-1] [I]; for (I = 0; I <num [2] [in]; I ++) {for (j = T; j> = pack [2] [start + I]. t; j --) {If (DP [C] [J-pack [2] [start + I]. t]! =-1) DP [C] [J] = DP [C] [J]> DP [C] [J-pack [2] [start + I]. t] + pack [2] [start + I]. h? DP [C] [J]: DP [C] [J-pack [2] [start + I]. t] + pack [2] [start + I]. h ;}} C ++ ;}int main () {int I = 0, j = 0, m = 0, S = 0, start = 0, ANS =-1; while (scanf ("% d", & N, & T )! = EOF) {c = 1; memset (index1, 0, sizeof (index1); memset (IND, 0, sizeof (IND); memset (DP,-1, sizeof (DP); for (I = 0; I <= T; I ++) DP [0] [I] = 0; for (I = 0; I <n; I ++) {scanf ("% d", & M, & S); num [s] [index1 [s] ++] = m; // index1 [s] indicates the number of classes in S for (j = 0; j <m; j ++, IND [s] ++) {scanf ("% d", & Pack [s] [ind [s]. t); scanf ("% d", & Pack [s] [ind [s]. h) ;}} start = 0; For (j = 0; j <index1 [0]; j ++) // The total number of groups of classes belonging to 0 {zeropack (J, start); start + = num [0] [J]; C ++;} start = 0; for (j = 0; j <index1 [1]; j ++) {onepack (J, start); start + = num [1] [J]; c ++;} start = 0; For (j = 0; j <index1 [2]; j ++) {twopack (J, start ); start + = num [2] [J];} printf ("% d \ n", DP [C-1] [T]);} return 0 ;}