HDU 1024 Max Sum Plus Plus (动态规划 最大M字段和)
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 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj(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(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj(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(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
Process to the end of file.
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
题目大意:有n的个数,需要你将他划分成m段互不相交的子段,求最大的子段和
思路:在这里向大家推荐一篇大佬的博客,讲解的清晰明了,让我获益匪浅 https://blog.csdn.net/winter2121/article/details/72848482 Orz
大佬的博客写的很清晰,而且附上图解很容易弄懂
dp[i][j] 代表前i个数并且以第i个数为末尾划分成了j段的最大和。
而dp[i][j] 有两种情况:1. 将第i个数划分到前面这一段取
2.将第i个数单独拿出来作为一段的开始
dp[i][j] = max(dp[i-1][j],max(前i个数中的最大和))+a[j];
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<algorithm> 3 #include<cstring> 4 5 using namespace std; 6 typedef long long LL; 7 const int INF = 0x3f3f3f3f; 8 const int maxn = 1000005; 9 LL n, m; 10 LL dp[maxn][2], a[maxn];//dp[i][j]代表以第i个数为结尾且分成j段的最大和 11 int main() 12 { 13 ios::sync_with_stdio(false); 14 while (/*cin>>m>>n*/scanf("%lld%lld", &m, &n) != EOF) { 15 for (int i = 1; i <= n; i++)scanf("%lld", &a[i]); 16 /*cin >> a[i];*/ 17 for (int i = 0; i <= n; i++)dp[i][0] = dp[i][1] = 0; 18 for (int i = 1, k = 1; i <= m; i++, k ^= 1) { 19 LL maxpre = -INF; dp[i - 1][k] = -INF;//避免与后面for循环的影响 20 for (int j = i; j <= n - m + i; j++) { 21 maxpre = max(maxpre, dp[j-1][k ^ 1]);//寻找上一行第i个数之前的最大值 22 dp[j][k] = max(dp[j - 1][k], maxpre) + a[j]; 23 } 24 } 25 LL ans = -INF; 26 for (int i = m; i <= n; i++) 27 ans = max(ans, dp[i][m&1]); 28 //cout << ans << endl; 29 printf("%lld\n", ans); 30 } 31 return 0; 32 }