題目鏈接:
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1832
題目思路:
題目的目的是使A得更多分。設某序列爲序列a,A取最優f(a),使其變成序列b,B取最優,一直取到序列爲0.
那麼怎麼取呢,取的這一步必定包含在從左端取一個到全部或者從右端取一個到全部取完或者不取。
即滿足關係式:
d[i][j] = max(d[i][j], Sum(i, j)-min(dp(i+k, j), dp(i, j-k)));(別人寫的,我隨便貼的)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#pragma warning(disable : 4996)
using namespace std;
const int N = 105;
int dp[N];
int sum[N];
int vis[N][N], d[N][N];
int n;
int f(int l, int r)
{
//第一次提交時間超限了,注意這裏可以使用記憶化搜索
if (vis[l][r]) return d[l][r];
vis[l][r] = 1;
int maxn = 0;
for (int i = l+1; i <= r; i++)
{
//第一次全部取完的情況沒有算上
maxn = min(maxn, f(i, r));
}
for (int i = l; i < r;i++)
maxn = min(maxn, f(l, i));
d[l][r] =sum[r]-sum[l]+dp[l]- maxn;
return d[l][r];
}
int main()
{
//freopen("in.txt", "r", stdin);
while (~scanf("%d", &n)&&n)
{
memset(sum, 0, sizeof(sum));
memset(vis, 0, sizeof(vis));
bool judge = false;
for (int i = 0; i < n; i++)
{
scanf("%d", &dp[i]);
if(i>0)
sum[i] = sum[i - 1] + dp[i];
else
sum[i] = dp[i];
if (dp[i] < 0)
judge = true;
}
if(!judge)
cout << sum[n - 1] << endl;
else
printf("%d\n", 2 * f(0, n - 1) - sum[n - 1]);
}
//system("pause");
return 0;
}