uva 10003 Cutting Sticks （动态规划：区间DP）

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.

题目意思  ： 有一根长度为l的木棍，木棍上面有m个切割点，每一次切割都要付出当前木棍长度的代价，问怎样切割有最小代价
思路：
方法一 区间DP
代码：
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <stack>
#include <queue>
#include <set>
using namespace std;
#define MAXN 55
int len , m;
int dp[MAXN][MAXN];
int cut[MAXN];//记录切割点

void solve(){
int i , j , k , p , tmp , min;
cut[0] = 0 ;
cut[m+1] = len; //增加切点，那么切点有0-m+1
memset(dp , 0 , sizeof(dp));
for(p = 1 ; p <= m+1 ; p++){//区间长度0-m+1
for(i = 0 ; i <= m+1-p ; i++){//区间起点0-m+1-p
j = i+p; min = 999999999;//j为终点
for(k = i+1 ; k < j ; k++){//枚举k ,注意是i+1
tmp = dp[i][k]+dp[k][j]+cut[j]-cut[i];//这个地方注意不是从k+1开始
if(tmp < min) min = tmp;//更新min
}
if(min != 999999999) dp[i][j] = min;//如果min有更新过则赋值给dp[i][j]
}
}
printf("The minimum cutting is %d.\n" , dp[0][m+1]);//答案就是0-m+1这个区间
}

int main(){
//freopen("input.txt" , "r" , stdin);
while(scanf("%d" , &len) && len){
scanf("%d",&m);
for(int i = 1;i <= m;i++)
scanf("%d",&cut[i]);
solve();
}
return 0;
}

方法二：记忆化搜索，对于一根长为len的木棍，从x处截断，则花费len，并把当前木棍截为（x,len-x）两段用记忆化搜索的方法就很好写dp[i][j] = min(dp[i][j], dp[i][k]+dp[k][j]+a[j]-a[i])
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#define maxn 1010
#define inf 0x7f7f7f7f
using namespace std;
int cut[maxn];
int dp[maxn][maxn];
int dfs(int i,int j){
if(dp[i][j]!=-1){
return dp[i][j];
}
if(i+1==j) return dp[i][j]=0;

dp[i][j]=inf;
int temp;
for(int k=i+1;k<j;k++) {
temp=dfs(i,k)+dfs(k,j)+cut[j]-cut[i];
dp[i][j]=min(dp[i][j],temp);
}
return dp[i][j];
}
int main(){
int n,len;
while(scanf("%d",&len) && len){
scanf("%d",&n);
cut[0]=0;
cut[n+1]=len;
for(int i=1;i<=n;i++) scanf("%d",&cut[i]);
memset(dp,-1,sizeof(dp));
printf("The minimum cutting is %d.\n",dfs(0,n+1));
}
return 0;
}