uva 10891

dp

状态转移方程:d(i,j)=sum(i,j)-min{d(i+1,j),d(i+2,j)...d(j,j),d(i,j-1),d(i,j-2)...d(i,i),0}
可化为d(i,j)=sum(i,j)-min{f(i+1,j),g(i,j-1),0}

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=100+10;
int arr[maxn];
int f[maxn][maxn],d[maxn][maxn],s[maxn],g[maxn][maxn];
int n;
int min(int a,int b)  { return a<b?a:b; }
int main()
{
	while(scanf("%d",&n)&&n)
	{
		int i,j,k;
		s[0]=0;
		for(i=1;i<=n;i++) { scanf("%d",&arr[i]); s[i]=s[i-1]+arr[i]; }
		for(i=1;i<=n;i++) f[i][i]=g[i][i]=d[i][i]=arr[i];
		for(i=2;i<=n;i++)
		{
			for(j=1;j+i-1<=n;j++)
			{
				k=j+i-1;
				int tem=min(f[j+1][k],g[j][k-1]);
				d[j][k]=s[k]-s[j-1]-min(tem,0);
				f[j][k]=min(d[j][k],f[j+1][k]);
				g[j][k]=min(d[j][k],g[j][k-1]);
			}
		}
		printf("%d\n",2*d[1][n]-s[n]);
	}
	return 0;
}


 

原文地址:https://www.cnblogs.com/lj030/p/3002220.html