Day9 - E - Max Sum Plus Plus HDU - 1024

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S  ₁, S  ₂, S  ₃, S  ₄ ... S  _x, ... S  _n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S  _x ≤ 32767). We define a function sum(i, j) = S  _i + ... + S  _j (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i  ₁, j  ₁) + sum(i  ₂, j  ₂) + sum(i  ₃, j  ₃) + ... + sum(i  _m, j  _m) maximal (i  _x ≤ i  _y ≤ j  _x or i  _x ≤ j  _y ≤ j  _x is not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i  _x, j  _x)(1 ≤ x ≤ m) instead. ^_^

InputEach test case will begin with two integers m and n, followed by n integers S ₁, S ₂, S ₃ ... S _n.
Process to the end of file.
OutputOutput the maximal summation described above in one line.
Sample Input

1 3 1 2 3
2 6 -1 4 -2 3 -2 3

Sample Output

6
8


        
 

Hint

Huge input, scanf and dynamic programming is recommended.

思路：最大m段子段和问题,设sum[n][m]为前n个数，m段子段和的最大值，状态转移有三:
1.a[n]不属于该第m段最大和，sum[n][m] = sum[n-1][m]
2.a[n]属于该第m段最大和，且为新的一段 sum[n][m] = sum[n-1][m-1] + a[n]
3.a[n]属于该第m段最大和，且不为新的一段 sum[n][m] = sum[n-1][m] + a[n]
由状态转移方程易知，1与3互相矛盾，就要寻找第二个状态转移方程来辅助计算，令b[n][m]为包含a[n]的前n个数的m段最大子段和，则sum[n][m]为前n个数的m段最大子段和，a[n]不必定包含在内，据此知，sum[n][m] = max(b[j][m])(m<=j<=n)，最终答案就是sum[n][m]
由上述分析，可得出b[n][m]的状态转移方程，a[n]属于第m段/不属于第m段，即b[n][m] = max(b[n-1][m], sum[n-1][m-1]) + a[n],经过该式能够得出计算顺序，先计算b[n][m],其中须要sum[n-1][m-1]与b[n-1][m],再更新sum[n][m]),而在计算sum[n][m] = max(b[j][m])(n<=j<=m),类比彻底背包的优化，在计算sum[n][m]时，sum[n][m] = max(b[j][m])(m<=j<=n) = max(b[n][m], sum[n-1][m]),由于sum[n-1][m] = max(b[x][m])(m<=x<=n-1)，等式成立,咱们枚举的顺序是从小到大，因此在sum[n][m]计算以前sum[n-1][m]已经计算好了，不须要再重复枚举了

const int maxm = 1005;

int sum[maxm][maxm], b[maxm][maxm], a[maxm];

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int n, m;
    int tmp;
    while(cin >> m >> n) {
        memset(sum, 0, sizeof(sum)), memset(a, 0, sizeof(a)), memset(b, 0, sizeof(b));
        for(int i = 1; i <= n; ++i) cin >> a[i];
        for(int i = 1; i <= m; ++i) {
            for(int j = i; j <= n; ++j) {
                b[j][i] = max(b[j-1][i], sum[j-1][i-1]) + a[j];
                sum[j][i] = max(sum[j-1][i], b[j][i]);
            }
        }
        cout << sum[n][m] << endl;
    }
    return 0;
}

可是注意，该题的范围是1e6,二维显然空间炸了，就须要优化为一维的滚动数组形式，咱们将计算的式子都列出来：
b[n][m] = max(b[n-1][m], sum[n-1][m-1]) + a[n], sum[n][m] = max(sum[n-1][m], b[n][m])
滚动时从小到大，该位置为j，则j以前的是[][m],j以后的是[][m-1](上一轮的)，则先计算b[],再更新sum[],由于sum和b有冲突，先sum[n-1][m-1],再sum[n-1][m],就用一个tmp来记录

const int maxm = 1e6+5;

int sum[maxm], b[maxm], a[maxm];

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int n, m;
    int tmp;
    while(cin >> m >> n) {
        memset(sum, 0, sizeof(sum)), memset(a, 0, sizeof(a)), memset(b, 0, sizeof(b));
        for(int i = 1; i <= n; ++i) cin >> a[i];
        for(int i = 1; i <= m; ++i) {
            tmp = -0x3f3f3f3f;
            for(int j = i; j <= n; ++j) {
                b[j] = max(b[j-1], sum[j-1]) + a[j]; // 此处sum[j-1] 是第i-1段
                sum[j-1] = tmp;  // 更新sum[j-1] 为第i段的, 该tmp是第i段的j-1,由上一次的循环继承而来
                tmp = max(sum[j-1], b[j]); // 更新tmp, 该tmp是第i段的j,为下一次循环中的的j-1
            }
            sum[n] = tmp;
        }
        cout << sum[n] << "\n";
    }
    return 0;
}

 参考微博：https://www.cnblogs.com/chuckcpc/p/dp_Max_M_Sum.html