紫书动规例题9-9 UVA

题目链接:

题意:

题解:

dp[i][j]:=切割小木棍i~j的最优费用，枚举中间切割点，这段的费用是该段的长度
d[i][j] = min(d[i][j],dp(i,k)+dp(k,j)+a[j]-a[i]);

代码:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define MS(a) memset(a,0,sizeof(a))
 5 #define MP make_pair
 6 #define PB push_back
 7 const int INF = 0x3f3f3f3f;
 8 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 9 inline ll read(){
10     ll x=0,f=1;char ch=getchar();
11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
13     return x*f;
14 }
15 //////////////////////////////////////////////////////////////////////////
16 const int maxn = 1e5+10;
17 
18 int len,n;
19 int a[maxn];
20 int d[100][100];
21 
22 int dp(int i,int j){
23     if(i+1 == j) return 0;
24     if(d[i][j] != INF) return d[i][j];
25 
26     for(int k=i+1; k<j; k++){
27         d[i][j] = min(d[i][j],dp(i,k)+dp(k,j)+a[j]-a[i]);
28     }
29     return d[i][j];
30 }
31 
32 int main(){
33     while(cin>>len && len){
34         cin >> n;
35         for(int i=1; i<=n; i++)
36             a[i] = read();
37         a[0] = 0; a[n+1] = len;
38 
39         // dp[i][j] = dp[i][k]+dp[k][j];
40         for(int i=0; i<=n+1; i++)
41             for(int j=0; j<=n+1; j++)
42                 d[i][j] = INF;
43         printf("The minimum cutting is %d.
", dp(0,n+1));
44     }
45     return 0;
46 }

紫书动规 例题9-9 UVA

题目链接:

题意:

题解:

代码:

紫书动规例题9-9 UVA