南阳理工OJ_题目1030 Yougth's Game[Ⅲ]

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//定义状态dp[i][j]为从i到j上先取数者的得分，那么后取数者的得分就是sum(i,j)-dp[i][j]//状态转移方程: //t1 = a[i]+(sum[j]-sum[i]-d[i+1][j]);取a[i],在剩余的i+1--j中,b先取,b取得的为d[i+1][j],//t2 = a[j]+(sum[j-1]-sum[i-1]-d[i][j-1]);sum[j]-sum[i]是i+1--j的和//d[i][j] = max(t1, t2);#include <iostream>#include <cstring>using namespace std;int dp();int n;int a[1010];int d[1010][1010];int sum[1010];int main(){    while(cin >> n)    {        sum[0] = 0;        for(int i = 1; i <= n; i++)        {            cin >> a[i];            sum[i] = sum[i-1] + a[i];        }        cout << dp() << '\n';    }}int dp(){    memset(d, 0, sizeof(d));    for(int l = 1; l <= n; l++)        for(int i = 1, j = l; j <= n; j++, i++)        {            int t1, t2;            t1 = a[i] + (sum[j] - sum[i] - d[i+1][j]);            t2 = a[j] + (sum[j-1] - sum[i-1] - d[i][j-1]);            d[i][j] = max(t1, t2);        }    return d[1][n] - (sum[n]-d[1][n]);}

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