题目描述
Farmer John is building a nicely-landscaped garden, and needs to move a large amount of dirt in the process.
The garden consists of a sequence of N flowerbeds (1 <= N <= 100), where flowerbed i initially contains A_i units of dirt. Farmer John would like to re-landscape the garden so that each flowerbed i instead contains B_i units of dirt. The A_i's and B_i's are all integers in the range 0..10.
To landscape the garden, Farmer John has several options: he can purchase one unit of dirt and place it in a flowerbed of his choice for X.HecanremoveoneunitofdirtfromaflowerbedofhischoiceandhaveitshippedawayforX.HecanremoveoneunitofdirtfromaflowerbedofhischoiceandhaveitshippedawayforY. He can also transport one unit of dirt from flowerbed i to flowerbed j at a cost of $Z times |i-j|. Please compute the minimum total cost for Farmer John to complete his landscaping project.
输入
* Line 1: Space-separated integers N, X, Y, and Z (0 <= X, Y, Z <= 1000).
* Lines 2..1+N: Line i+1 contains the space-separated integers A_i and B_i.
输出
* Line 1: A single integer giving the minimum cost for Farmer John's landscaping project.
样例输入
样例输出
题解：这道题目由于数据比较小，所以可以将数据来张开处理。随后，用动态规划解决。
#include <cstdio>
#include <cmath>
int n,x,y,z,la,lb,a[1005],b[1005],f[1005][1005];
using namespace std;
inline int minn(int x,int y)
{
return x<y?x:y;
}
int main()
{
scanf("%d%d%d%d",&n,&x,&y,&z);
for (int i=1;i<=n;i++)
{
int xx,yy;
scanf("%d%d",&xx,&yy);
for (int j=1;j<=xx;j++) a[++la]=i;//展开初始堆
for (int j=1;j<=yy;j++) b[++lb]=i;//展开目标堆
}
for (int i=1;i<=la;i++) f[i][0]=i*y;//减少i个泥土，要用i*y的费用
for (int i=1;i<=lb;i++) f[0][i]=i*x;//增加i个泥土，要用i*x的费用
for (int i=1;i<=la;i++)
for (int j=1;j<=lb;j++)
f[i][j]=minn(f[i-1][j]+y,minn(f[i][j-1]+x,f[i-1][j-1]+z*abs(a[i]-b[j])));//算最小值
printf("%d\n",f[la][lb]);
return 0;
}