UVA 10003 —— 区间DP
10003 - Cutting Sticks
Time limit: 3.000 seconds
| Cutting Sticks |
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.
2000-08-21 题意是给你一个n长度的木棒，然后从m个位置进行分割，每次每个的花费为当前木棒的长度，让你安排这些分割的次序使之花费最小。
思路就是枚举区间中的一个值进行区间DP。
/*
ID: xinming2
PROG: stall4
LANG: C++
*/
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 15 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
const int mod = (int)1e9 + 7;
typedef long long LL;
const LL MOD = 1000000007LL;
const double PI = acos(-1.0);
typedef pair<int , int> pi;
#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
int dp[MAXN][MAXN] ;///dp[i][j]表示从i到j的最小花费
int a[MAXN] , vis[MAXN][MAXN];
int DP(int i , int j)
{
if(vis[i][j])return dp[i][j];
vis[i][j] = 1;
if(j - i == 1)return 0;
int min_i = INF;
for(int k = i + 1 ; k < j ; k++)
{
min_i = min(min_i , DP(i , k) + DP(k , j) + a[j] - a[i]);
}
dp[i][j] =min_i;
return min_i;
}
int main()
{
int n , m;
while(~scanf("%d" , &n) , n)
{
clr(dp , 0);
clr(vis , 0);
scanf("%d" , &m);
for(int i = 1 ;i <= m ; i++)
{
scanf("%d", &a[i]);
}
a[++m] = n;
printf("The minimum cutting is %d.\n" , DP(0 , m));
}
return 0;
}