HDU 1024 Max Sum Plus Plus 动态规划

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1024

题目大意：n个数分成两两不相交的m段，求使这m段和的最大值。

解题思路：比较坑的点：n² 能过；long long超时，int AC。 

dp[i][j]:= 在选择第i个数的情况下前i个数分成j段的最大值
dp[i][j] = max(dp[i - 1][j] + a[i], max(dp[x][j - 1] -> dp[x][j - 1]) + a[i]) x < i

由于n<1000000，并且更新dp[i][j]时只用到了j和j-1的部分，因此采用滚动数组记录选择第i个人时前i个人分成j段的和最大值；同时，max(dp[x][j - 1] -> dp[x][j - 1]) 可以在更新dp[i]的同时记录，这样就将时间复杂度降低到了n² （所以为什么n2能过？？？）

另外就是一些细节问题。

代码：

const int inf = 0x3f3f3f3f;
//dp[i][j]:= 在选择第i个数的情况下前i个数分成j段的最大值
//dp[i][j] = max(dp[i - 1][j] + a[i], max(dp[x][j - 1] -> dp[x][j - 1]) + a[i]) x < i
const int maxn = 1e6 + 5;
int dp[maxn];
int pre[maxn];
int a[maxn], n, m;

int solve(){
    memset(dp, 0, sizeof(dp));
    memset(pre, 0, sizeof(pre));
    int tmax = -inf;
    for(int j = 1; j <= m; j++){
        tmax = -inf;//记录前i个人本次的最大值 
        for(int i = j; i <= n; i++){
            dp[i] = max(dp[i - 1] + a[i], pre[i - 1] + a[i]);//使用的是未更新的值，对应j-1的 
            pre[i - 1] = tmax;    //再更新 
            tmax = max(tmax, dp[i]);
        }
    }
    return tmax; //不一定是dp[n]，最后一个可能不用选 
}

int main(){
    while(scanf("%d %d", &m, &n) != EOF){
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        int ans = solve();
        printf("%d\n", ans);
    }
}

题目：

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30900    Accepted Submission(s): 10889

Problem Description

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 S₁, S₂, S₃, S₄ ... 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(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + 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. ^_^

 

 

Input

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
Process to the end of file.

 

 

Output

Output 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.

 

 

Author

JGShining（极光炫影）