Description
You have to cut a wood stick into pieces. The most affordable company, The Analog Cutting Machinery, Inc. (ACM), charges money according to the length of the stick being cut. Their procedure of work requires that they only make one cut at a time.
It is easy to notice that different selections in the order of cutting can led to different prices. For example, consider a stick of length 10 meters that has to be cut at 2, 4 and 7 meters from one end. There are several choices. One can be cutting first at 2, then at 4, then at 7. This leads to a price of 10 + 8 + 6 = 24 because the first stick was of 10 meters, the resulting of 8 and the last one of 6. Another choice could be cutting at 4, then at 2, then at 7. This would lead to a price of 10 + 4 + 6 = 20, which is a better price.
Your boss trusts your computer abilities to find out the minimum cost for cutting a given stick.
The next line consists of n positive numbers ci ( 0 <ci < l) representing the places where the cuts have to be done, given in strictly increasing order.
An input case with l = 0 will represent the end of the input.
100 3 25 50 75 10 4 4 5 7 8 0
The minimum cutting is 200. The minimum cutting is 22.
之所以要加上a[j]-a[i],是要加上当前切割的决策的费用。
注意初始化问题,dp[i][i+1]=0,
也就是这些本身就是现成的小段,不能继续分割,决策费用也为0,总费用为0。
#include <iostream>
#include <string.h>
using namespace std;
int dp[55][55],c[55];
int DFS(int x,int y)
{
if(dp[x][y]!=-1)
return dp[x][y];
int mins=9999999999,i;
for(i=x+1;i<y;i++)
{
int temp=DFS(x,i)+DFS(i,y)+c[y]-c[x];
if(temp<mins)
mins=temp;
}
return (dp[x][y]=mins);
}
int main()
{
int len,n,i;
while(cin>>len&&len!=0)
{
cin>>n;
for(i=1;i<=n;i++)
cin>>c[i];
c[0]=0;
c[n+1]=len;
memset(dp,-1,sizeof(dp));
for(i=0;i<=n;i++)
dp[i][i+1]=0;
cout<<"The minimum cutting is "<<DFS(0,n+1)<<"."<<endl;
}
return 0;
}