HDU_1158 Employment Planning(DP)

　　又是一道纠结的dp题，做dp快做的崩溃了。。。

 状态表示:    

 Dp[i][j]为前i个月的留j个人的最优解;Num[i]<=j<=Max{Num[i]};
                j>Max{Num[i]}之后无意义,无谓的浪费 记Max_n=Max{Num[i]};
    Dp[i-1]中的每一项都可能影响到Dp[i],即使Num[i-1]<<Num[i]
    所以利用Dp[i-1]中的所有项去求Dp[i];
    对于Num[i]<=k<=Max_n,   

 当k<j时, 招聘;  当k>j时, 解雇  然后求出最小值
    Dp[i][j]=min{Dp[i-1][k…Max_n]+(招聘,解雇,工资);   

代码：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N = 10000;
const int inf = 0xfffffff;

int mon[13];
int dp[13][N];

int main()
{
//freopen("data.in", "r", stdin);

int m, i, j, k;
while(scanf("%d", &m), m)
    {
int hire, fire, salary, max, min, cost;

        memset(dp, 0, sizeof(dp));
        memset(mon, 0, sizeof(mon));

        scanf("%d%d%d", &hire, &salary, &fire);
        max = -inf;

for(i = 1; i <= m; i++)
        {
            scanf("%d", mon + i);
            max = max > mon[i] ? max : mon[i];
        }

for(i = mon[1]; i <= max; i++)
            dp[1][i] = (hire+salary) * i;

for(i = 2; i <= m; i++)
        {
for(j = mon[i]; j <= max; j++)
            {
                min = inf;
for(k = mon[i-1]; k <= max; k++)
                {
if(k <= j)
                        cost = (j-k)*hire + j*salary + dp[i-1][k];
else
                        cost = (k-j)*fire + j*salary + dp[i-1][k];
if(min > cost)
                        min = cost;
                }
//printf("%d\n", min);
                dp[i][j] = min;
            }
        }

        min = inf;
for(i = mon[m]; i <= max; i++)
if(min > dp[m][i])
                min = dp[m][i];
        printf("%d\n", min);
    }
return 0;
}