UVa-10003 - Cutting Sticks

区间dp，O(n^3)的算法很容易想。先递增枚举区间的长度d，再枚举起点i，再枚举切点k(i<k<j)(j=i+d))。代码如下：(0.039s)

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn=55;
int l,n;
int p[maxn];
int dp[maxn][maxn];
int main()
{
    while(scanf("%d",&l)&&l)
    {
        memset(dp,0x3f,sizeof(dp));
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&p[i]);
        p[n+1]=l;
        for(int d=1;d<=n+1;d++)
            for(int i=0;i+d<=n+1;i++)
            {
                if(d==1) dp[i][i+1]=0;
                else
                {
                    for(int k=i+1;k<i+d;k++)
                        dp[i][i+d]=min(dp[i][i+d],dp[i][k]+dp[k][i+d]);
                    dp[i][i+d]+=p[i+d]-p[i];
                }
            }
        printf("The minimum cutting is %d.\n",dp[0][n+1]);
    }
}

然后看书上说可以用四边形不等式优化到O(n^2)，今天新买的黑书到手了，看了看讲解(p151,最优二分检索树)。大意是找k值的决策有单调性，利用此单调性并记录下来，可以在O(1)找到决策。代码如下：(0.029s) (rank 4!)

关于四边形不等式，http://blog.csdn.net/shiwei408/article/details/8791011 可以看这个博客。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn=55;
int l,n;
int p[maxn];
int dp[maxn][maxn];
int kp[maxn][maxn];
int main()
{
    while(scanf("%d",&l)&&l)
    {
        memset(dp,0x3f,sizeof(dp));
        memset(kp,0,sizeof(kp));
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&p[i]);
        p[n+1]=l;
        for(int d=1;d<=n+1;d++)
            for(int i=0;i+d<=n+1;i++)
            {
                int j=i+d;
                if(d==1)
                {
                    dp[i][i+1]=0;
                    kp[i][i+1]=i+1;
                }
                else
                {
                    for(int k=kp[i][j-1];k<=kp[i+1][j];k++)
                        if(dp[i][k]+dp[k][j]<dp[i][j])
                        {
                            dp[i][j]=dp[i][k]+dp[k][j];
                            kp[i][j]=k;
                        }
                    dp[i][j]+=p[j]-p[i];
                }
            }
        printf("The minimum cutting is %d.\n",dp[0][n+1]);
    }
}