HDU 1158 - Employment Planning
传送门
HDU 1158 - Employment Planning
题意
给出雇佣,解雇,每个月所要的花费,再给出每个月最少要有的人,求总的费用最小。
思路
$dp[i][j] = min(dp[i][j], dp[i][k] + (j - k) * hire + salary * j)$
k表示上个月的雇佣人数,如果k>j,说明有人被解雇了,就要换成解雇的费用,枚举k即可。
代码
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MP(a, b) make_pair(a, b)
using namespace std;
const int MAXN = 10000;
const int INF = 0x3f3f3f3f;
int dp[13][MAXN], pp[13];
int main()
{
//freopen("input.txt", "r", stdin);
int n, i, j, hire, salary, fire, nmax;
while (scanf("%d", &n), n)
{
memset(dp, INF, sizeof dp);
nmax = -1;
scanf("%d%d%d", &hire, &salary, &fire);
for (i = 1; i <= n; i++)
{
scanf("%d", &pp[i]);
nmax = max(nmax, pp[i]);
}
for (i = pp[1]; i <= nmax; i++)
dp[1][i] = i * (hire + salary);
for (i = 2; i <= n; i++)
{
for (j = pp[i]; j <= nmax; j++)
for (int k = pp[i - 1]; k <= nmax; k++)
{
if (j > k)
dp[i][j] = min(dp[i][j], dp[i - 1][k] + (j - k) * hire + j * salary);
else
dp[i][j] = min(dp[i][j], dp[i - 1][k] + (k - j) * fire + j * salary);
}
}
int ans = INF;
for (i = pp[n]; i <= nmax; i++)
ans = min(ans, dp[n][i]);
printf("%d\n", ans);
}
return 0;
}