Poj 1384 piggy-Bank (full backpack)

Poj 1384 piggy-Bank (full backpack)

Http://poj.org/problem? Id = 1384

Question:

There are now N types of coins, each of which has a specific weight cost [I] grams and its corresponding value Val [I]. each coin can be used infinitely. it is known that the total weight of all coins in a piggy bank is exactly M grams. How many coins are the most valuable in this piggy bank? If M grams cannot exist, output "This is impossible .".

Analysis:

Because each coin can be used infinitely, it is obviously a complete backpack problem.

This questionRestrictions: The total weight of a coin is equal to M.

This questionTarget Condition: The total value of coins should be as small as possible.

So that DP [I] [J] = X indicates that only the first I currency is used, and the minimum value when the total weight reaches J grams is X.

Initialization: All DP values are Inf (infinite) and DP [0] = 0. (because the goal of this question is to minimize the value, the initialization is infinite. to maximize the target, initialize it to-1 .)

State Transfer: DP [I] [J] = min (DP [I-1] [J], DP [I] [J-cost [I] + val [I])

In the above equation, the former indicates that none of the I-th coins are selected, and the latter indicates that at least one I-th coin is selected.

The final request is DP [N] [M]. (If DP [N] [m] = inf, it indicates that M grams are inaccessibility.)

This project uses a rolling array, so DP only has the [J] dimension.

AC code:

# Include <cstdio> # include <cstring> # include <algorithm> using namespace STD; const int maxn = 10000 + 5; # define INF 1e9int N; // currency type int m; // maximum weight int DP [maxn]; int cost [maxn]; // int Val [maxn] for each currency weight; // int main () {int t for each currency value; scanf ("% d", & T); While (t --) {// read data int M1, M2; scanf ("% d", & M1, & m2, & N); M = m2-m1; For (INT I = 1; I <= N; I ++) scanf ("% d ", & Val [I], & cost [I]); // initialize for (INT I = 0; I <= m; I ++) DP [I] = inf; DP [0] = 0; // recursive process for (INT I = 1; I <= N; I ++) {for (Int J = cost [I]; j <= m; j ++) DP [J] = min (DP [J], DP [J-cost [I] + val [I]);} // output result if (DP [m] = inf) printf ("This is impossible. \ n "); else printf (" the minimum amount of money in the piggy-bank is % d. \ n ", DP [m]);} return 0 ;}

Poj 1384 piggy-Bank (full backpack)