Bzoj 4145 [AMPPZ2014] The Prices Problem solving report

The feeling is also a small fresh question.

We considered setting the status $Dp [I][s]$ said that after considering the previous $i $ store, the purchase status was the minimum cost of $s $.

If you move, enumerate each store $i $, and first make:

$ $Dp [I][s] = Dp[i-1][s] + d[i]$$

This process means arriving at the store.

Then enumerate each state $s $, and then enumerate each item that is not in the $s $ $j $, so that:

$ $Dp [I][s + \{j\}] = min (Dp[i][s + \{j\}], Dp[i][s] + cost[i][j]) $$

This process is equivalent to a 01 backpack.

Finally make $Dp [i][s] = min (Dp[i][s], Dp[i-1][s]) $ to see if the purchase plan is cost-effective at the store $i $.

The Complete collection is $S $, so the final answer is $Dp [n][s]$.

Time complexity $O (nm2^m) $, spatial complexity $O (N2^M) $.

1#include <cstdio>2 #defineMin (A, B) ((a) < (b)? (a): (b))3 #defineN 100 + 54 #defineM 16 + 55 #defineSIZE 1 << 166 #defineINF 5931196817 8 intN, M, W[n], map[n][m], dp[n][size];9 Ten intMain () One { Ascanf"%d%d", &n, &m); -      for(inti =1; I <= N; i + +) -     { thescanf"%d", W +i); -          for(intj =1; J <= M; J + +) -scanf"%d", Map[i] +j); -     } +      for(ints =0; S < (1<< m); S + +) -dp[0][s] =INF; +dp[0][0] =0; A      for(inti =1; I <= N; i + +) at     { -          for(ints =0; S < (1<< m); S + +) -Dp[i][s] = dp[i-1][s] +W[i]; -          for(intj =1; J <= M; J + +) -              for(ints =0; S < (1<< m); S + +) -                 if((S & (1<< J-1)) ==0) inDp[i][s ^ (1<< J-1)] = min (dp[i][s ^ (1<< J-1)], Dp[i][s] +map[i][j]); -          for(ints =0; S < (1<< m); S + +) toDp[i][s] = min (Dp[i][s], dp[i-1][s]); +     } -printf"%d\n", dp[n][(1<< m)-1]); the      *     return 0; $}

4145_gromah

Bzoj 4145 [AMPPZ2014] The Prices Problem solving report