uva 10896 DP

dp【i】【j】表示先手所能获得的最大值

dp【i】【j】 = sum【i】【j】 - min{ dp【i】【k】 , dp【q】【j】，0 }   k从i到j-1   q从i+1到j

AC代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

int dp[110][110];
int f[110][110], g[110][110];
int N, num[110], sum[110];

int main(){
    while( cin >> N && N ){
        sum[0] = 0;
        for( int i = 1; i <= N; i++ ){
            cin >> num[i];
            sum[i] = num[i] + sum[i-1];
        }
        for( int i = 1; i <= N; i++ ){
            dp[i][i] = f[i][i] = g[i][i] = num[i];
        }
        for( int L = 1; L < N; L++ ){
            for( int i = 1; i + L <= N; i++ ){
                int j = i + L;
                int temp = 0;
                temp = min( temp, f[i+1][j] );
                temp = min( temp, g[i][j-1] );
                dp[i][j] = sum[j] - sum[i-1] - temp;
                f[i][j] = min( dp[i][j], f[i+1][j] );
                g[i][j] = min( dp[i][j], g[i][j-1] );
            }
        }
        cout << 2 * dp[1][N] - sum[N] << endl;
    }
    return 0;
}