uva 10891 Game of Sum 区间dp

题意：有一个长度为n的整数序列，A和B轮流取左端或者右端部分，使自己得分高。

整数序列的值是一定的，所以使一个人得分越高，即是另一个人的得分越低。d(i,j)表示原序列i~j个，则d(i,j) = sum(i,j) - min{d(i+1,j),d(i+2,j)....d(j,j),d(i,j-1)...d(i,i), 0}，0 表示全部取完，若直接记忆化搜索的话，O(n^3)，若用递推，可以用f(i,j) = min{d(i,j),d(i+1,j)... d(j,j)}. g(i,j) = min(d(i,j),d(i,j-1)...d(i,i)}，则时间复杂度为O(n^2)

/*version1 O(n^3)*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define mem(name,value) memset(name,value,sizeof(name))
#define FOR(i,n) for(int i=1;i<=n;i++)
using namespace std;
const int maxn =100+10;
int sum[maxn],A[maxn],d[maxn][maxn],vis[maxn][maxn];
int dp(int l,int r){
    if(vis[l][r]) return d[l][r];
    int m = 0;
    for(int i=l+1;i<=r;i++) m = min(m,dp(i,r));
    for(int i=r-1;i>=l;i--) m = min(m,dp(l,i));
    d[l][r] = sum[r] - sum[l-1] - m;
    vis[l][r] = true;
    return d[l][r];
}
int main(){
    int n;
    while(scanf("%d",&n)==1 && n){
        sum[0] - 0;
        mem(vis,false);
        for(int i=1;i<=n;i++){
            scanf("%d",&A[i]);
            sum[i] = sum[i-1]+A[i];
        }
        int ans = dp(1,n);
        ans = 2*ans - sum[n];
        printf("%d\n",ans);
    }
    return 0;
}
/*version2 O(n^2)*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define mem(name,value) memset(name,value,sizeof(name))
#define FOR(i,n) for(int i=1;i<=n;i++)
using namespace std;
const int maxn = 100+10;
int sum[maxn],A[maxn];
int d[maxn][maxn],f[maxn][maxn],g[maxn][maxn];
int main(){
    int n;
    while(scanf("%d",&n)==1 && n){
        for(int i=1;i<=n;i++){
            scanf("%d",&A[i]);
            sum[i] = sum[i-1]+A[i];
        }
        for(int i=1;i<=n;i++) d[i][i] = f[i][i] = g[i][i] = A[i];
        for(int L=1;L<n;L++){
            for(int i=1;i+L<=n;i++){
                int j = i+L;
                int m = 0;
                m = min(m,f[i+1][j]);
                m = min(m,g[i][j-1]);
                d[i][j] = sum[j] - sum[i-1] - m;
                f[i][j] = min(d[i][j],f[i+1][j]);
                g[i][j] = min(d[i][j],g[i][j-1]);
            }
        }
        printf("%d\n",2*d[1][n]-sum[n]);
    }
    return 0;
}