Va-10163 storage keepers (01 backpack)

(A 10163 storage keepers (01 backpack)

Now there are m shopkeepers and N stores, each of which has a capacity value, and then a guardian can keep K stores, the security of these stores is PI/K. The store's security depends on the one with the lowest security level provided by the person guarding it. The maximum security of these stores depends on the store with the lowest security. I am asking how to hire these people to make the store the most secure, with the lowest cost.

Solution:

With the highest security requirements, it is necessary to determine whether every store owner needs to hire it and keep it secure after hiring it ). DP [I] [J] indicates the maximum security value of the front I stores when the front J daemon has been guarded. DP [I] [J] = max (DP [I] [J-1], min (DP [k] [J-1], PJ/K )) k> = 0 & K <I. you can save the J dimension and convert it into a rolling array.

Then, the highest security of these stores comes out of ANS, which requires the least cost. When the security of each store is maintained, the maximum number of stores that employ this person can guard is PI/ANS. If there are more stores, the security will not meet the requirements, then we thought the same way as above: only DP [I] [J] is the lowest cost for the front I store to use the front J daemon (to ensure the highest security ). DP [I] [J] = min (DP [I] [J-1], DP [k] [J-1] + PJ ).

If security is required, then K has corresponding requirements.

Code:

#include <cstdio>#include <cstring>#include <algorithm>#include <functional>using namespace std;const int N = 1005;const int M = 35;const int INF = 3000000;int p[M];int dp[N];int y[N];int Min (const int a, const int b) {return a < b ? a: b; }int Max (const int a, const int b) {return a > b ? a: b; }int main () {int n, m;while (scanf ("%d%d", &n, &m) , n || m) { for (int i = 1; i <= m; i++)scanf ("%d", &p[i]);sort (p + 1, p + 1 + m, greater<int>());memset (dp, 0, sizeof (dp));dp[0] = INF;for (int i = 1; i <= m; i++) for (int j = n; j >= 1; j--) {for (int k = 1; k <= j; k++) {dp[j] = Max (dp[j], Min (dp[j - k] , p[i] / k));}}int ans = dp[n];if (ans) {for (int i = 1; i <= n; i++)dp[i] = INF;dp[0] = 0;for (int i = 1; i <= m; i++)for (int j = n; j >= 1; j--) for (int k = Min(p[i]/ans, j); k >= 1; k--) {dp[j] = Min (dp[j], dp[j - k] + p[i]);}} elsedp[n] = 0;printf ("%d %d\n", ans, dp[n]);}return 0;}