uva 10891 Game of Sum (DP水题)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1832

　　博弈类的DP，通过枚举较短一段的最大得分推出较长一段的最大得分。转移方程是 dp[i][j] = sum[i][j] - min(dp[i][i], dp[i][i + 1], ... , dp[i][j - 1], dp[i + 1][j], dp[i + 2][j], ... ,dp[j][j])。

O(n^3)算法：

#define REP(i, n) for (int i = 0; i < (n); i++)
#define REP_1(i, n) for (int i = 1; i <= (n); i++)

int dp[N][N], sum[N];

int main() {
    int n;
    while (~scanf("%d", &n) && n) {
        sum[0] = 0;
        REP_1(i, n) {
            scanf("%d", &sum[i]);
            sum[i] += sum[i - 1];
        }
        _clr(dp);
        REP_1(d, n) {
            REP_1(i, n - d + 1) {
                int j = i + d - 1, tmp = sum[j] - sum[i - 1];
                dp[i][j] = tmp;
                REP(l, d - 1) {
                    dp[i][j] = max(dp[i][j], tmp - dp[i][i + l]);
                    dp[i][j] = max(dp[i][j], tmp - dp[j - l][j]);
                }
            }
        }
//        printf("%d %d\n", dp[1][n], sum[n] - dp[1][n]);
        printf("%d\n", (dp[1][n] << 1) - sum[n]);
    }
    return 0;
}

在计算较短段的时候记录头尾的最小值，可以得到O(n^2)算法：

#define REP(i, n) for (int i = 0; i < (n); i++)
#define REP_1(i, n) for (int i = 1; i <= (n); i++)

int dp[N][N], sum[N], rec[2][N];

int main() {
    int n;
    while (~scanf("%d", &n) && n) {
        sum[0] = 0;
        REP_1(i, n) {
            scanf("%d", &sum[i]);
            sum[i] += sum[i - 1];
        }
        _clr(dp);
        REP(i, 2) REP_1(j, n) rec[i][j] = inf;
        REP_1(d, n) {
            REP_1(i, n - d + 1) {
                int j = i + d - 1, tmp = sum[j] - sum[i - 1];
                dp[i][j] = max(tmp, max(tmp - rec[0][i], tmp - rec[1][j]));
                rec[0][i] = min(dp[i][j], rec[0][i]);
                rec[1][j] = min(dp[i][j], rec[1][j]);
            }
        }
//        printf("%d %d\n", dp[1][n], sum[n] - dp[1][n]);
        printf("%d\n", (dp[1][n] << 1) - sum[n]);
    }
    return 0;
}

——written by Lyon