Employment Planning
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2110 Accepted Submission(s): 832
Problem Description
A project manager wants to determine the number of the workers needed in every month. He does know the minimal number of the workers needed in each month. When he hires or fires a worker, there will be some extra cost. Once a worker is hired, he will get the salary even if he is not working. The manager knows the costs of hiring a worker, firing a worker, and the salary of a worker. Then the manager will confront such a problem: how many workers he will hire or fire each month in order to keep the lowest total cost of the project.
Input
The input may contain several data sets. Each data set contains three lines. First line contains the months of the project planed to use which is no more than 12. The second line contains the cost of hiring a worker, the amount of the salary, the cost of firing a worker. The third line contains several numbers, which represent the minimal number of the workers needed each month. The input is terminated by line containing a single '0'.
Output
The output contains one line. The minimal total cost of the project.
Sample Input
3
4 5 6
10 9 11
0
Sample Output
199
dp[i][j]表示前i个月最后一个月的总人数为j所花的最小费用
状态移动方程:dp[i][j] = min{dp[i-1][k] + cost[i][j]},其中cost[i][j]是第i月的花费,
1~当k<=j时，第i个月请了人所以cost[i][j] = j*salary + (j-k)*hire
2~当k>j时，第i个月炒了人，所以cost[i][j] = j*salary + (k-j)*fire
输入时记录了最多需要的人数。
因为他给的是每个月最少需要的人数，所以for(该月需要的最少人数——max_people)而不是for(1——该月需要的最少人数)
直接初始化第一个月。
View Code
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cctype>
#include <cstring>
#include <sstream>
#include <fstream>
#include <cstdlib>
#include <cassert>
#include <iostream>
#include <algorithm>
using namespace std;
//Constant Declaration
/*--------------------------*/
//#define LL long long
#define LL __int64
const int M=10000;
const int INF=1<<30;
const double EPS = 1e-11;
const double PI = acos(-1.0);
/*--------------------------*/
// some essential funtion
/*----------------------------------*/
void Swap(int &a,int &b){ int t=a;a=b;b=t; }
int Max(int a,int b){ return a>b?a:b; }
int Min(int a,int b){ return a<b?a:b; }
int Gcd(int a,int b){ while(b){b ^= a ^=b ^= a %= b;} return a; }
/*----------------------------------*/
//for (i = 0; i < n; i++)
/*----------------------------------*/
int dp[15][M];
int people[15];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
//int t, case1 = 0;
//scanf("%d", &t);
int n, m;
int i, j, k;
while (~scanf("%d", &n), n)
{
int hire, fire, salary;
scanf("%d%d%d", &hire, &salary, &fire);
int max_people = 0;
for (i = 1; i <= n; i++)
{
scanf("%d", &people[i]);
if (max_people < people[i])
{
max_people = people[i];
}
}
//memset(dp, 0, sizeof(dp));不需要
for (i = people[1]; i <= max_people; i++)//初始化第一个月。
{
dp[1][i] = i*salary + i*hire;
}
for (i = 2; i <= n; i++)
{
for (j = people[i]; j <= max_people; j++)
{
int minn = INF;//有了这个前面就不需要用O（n^2）初始化dp了。
for (k = people[i-1]; k<= max_people; k++)
{
if (minn > dp[i-1][k] + (j>=k?(j*salary+hire*(j-k)):(j*salary+fire*(k-j))))
{
minn = dp[i-1][k] + (j>=k?(j*salary+hire*(j-k)):(j*salary+fire*(k-j)));
}
}
dp[i][j] = minn;
}
}
int min = INF;
for (j = people[n]; j <= max_people; j++)
{
if (min > dp[n][j])
{
min = dp[n][j];
}
}
printf("%d\n", min);
}
return 0;
}