uva10003 - Cutting Sticks

Cutting Sticks

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.

Input 

The input will consist of several input cases. The first line of each test case will contain a positive number l that
represents the length of the stick to be cut. You can assume l < 1000. The next line will contain the number n (n <
50) of cuts to be made.
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.

Output 

You have to print the cost of the optimal solution of the cutting problem, that is the minimum cost of cutting the given stick. Format the output as shown below.

Sample Input 

100
3
25 50 75
10
4
4 5 7 8
0

Sample Output 

The minimum cutting is 200.
The minimum cutting is 22.

这题可以逆向思考，正向切割与反向合并一一对应，这题就变成了区间合并，有点类似矩阵链乘了。
枚举区间长度，从2开始一直到n+1，初始化区间长度为1的值为0.

递推代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int a[60],dp[60][60];
const int inf=1<<30;
int main()
{
int n,len;
while(scanf("%d",&len),len){
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",a+i);
a[0]=0,a[n+1]=len;
for(int l=2;l<=n+1;l++)
for(int i=0;i+l<=n+1;i++){
dp[i][i+l]=inf;
for(int k=i+1;k<i+l;k++)
dp[i][i+l]=min(dp[i][i+l],dp[i][k]+dp[k][i+l]+a[i+l]-a[i]);
}
printf("The minimum cutting is %d.\n",dp[0][n+1]);
}
return 0;
}

记忆化搜索代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int n,len;
int a[60],dp[1010][1010];
const int inf=1<<30;
int dfs(int s,int e){
int &ans=dp[s][e];
bool flag=true;
if(ans!=inf) return ans;
for(int i=1;i<=n;i++)
if(a[i]>s&&a[i]<e){
ans=min(ans,dfs(s,a[i])+dfs(a[i],e)+e-s);
flag=false;
}
if(flag) return ans=0;
return ans;
}
int main()
{
while(scanf("%d",&len),len){
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",a+i);
for(int i=0;i<=len;i++)
for(int j=0;j<=len;j++)
dp[i][j]=inf;
printf("The minimum cutting is %d.\n",dfs(0,len));
}
}

暴力枚举区间的，因此数组要开len*len，并且复杂度是O（len*len）
另一个版本的，也就是模仿递推的，相当于离散化过的。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int n,len;
int a[60],dp[60][60];
const int inf=1<<30;
int dfs(int s,int e){
int &ans=dp[s][e];
if(s+1==e) return ans=0;
if(ans!=inf) return ans;
for(int i=s+1;i<e;i++)
ans=min(ans,dfs(s,i)+dfs(i,e)+a[e]-a[s]);
return ans;
}
int main()
{
while(scanf("%d",&len),len){
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",a+i);
a[0]=0,a[n+1]=len;
for(int i=0;i<=n+1;i++)
for(int j=0;j<=n+1;j++)
dp[i][j]=inf;
printf("The minimum cutting is %d.\n",dfs(0,n+1));
}
}

这个空间和时间复杂度一下降到了O（n*n），是个不错的优化。