Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26474    Accepted Submission(s): 9201 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 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. ^_^   Input Each 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.   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）   思路： dp[i][j]表示分成i段前j个元素的最大和，同时这个最大和是由a[j]结尾的。 状态为i、j，值为dp[i][j]。 若a[j]与前面一段直接相连成为共同的一段，那么前面应该已经有了i段： dp[i][j] = {dp[i][j-1] + a[j]} 若a[j]单独成一段，那么前面应该是已经有了i-1段： dp[i][j] = {dp[i-1][k] + a[j]} (i-1<=k<j) 所以状态转移方程为 dp[i][j] = {max(dp[i][j-1], dp[i-1][k]) + a[j]} (i-1<=j<k) 有了状态转移方程，剩下的就是优化了，把状态转移表画出来。 首先对一些特殊情况做一些处理，定义出i，j的边界条件：     1、第i段可以以任何一个数字结尾；     2、某段以a[j]结尾的时候，表示前j个元素以j结尾的最大和，也就是说，最多只能分成j段(现在一共只有j个元素，最多的情况下就是每个数字单独成一段，那么刚好是j段)，所以无论在什么情况下总有i<=j。 这里以题目中的 -1 4 -2 3 -2 3为例，状态转移表填写过程如下： 填写对角线元素的方法是，dp[i][j] = {dp[i-1][j-1] + a[j]} (i=j) 然后根据状态转移方程接着填 注意到，这里填每一行的时候，其状态仅与上一行有关。也就是说，仅仅依靠第i-1行就能够得到完整的第i行，同样，仅仅根据第i行就能填出i+1行，可以用滚动数组来优化空间，下面代码就用一个dp[2][M]大小的数组来完成整个表的填写。   #include <iostream> #include <cstring> #include <cstdio> #define M 1000010 #define INF 9999999 using namespace std; int a[M]; int dp[2][M]; int Max(int a, int b) { return a>b?a:b; } int main() { int m, n; while (scanf("%d %d", &m, &n)!=EOF) { for (int i=1; i<=n; i++) { scanf("%d", &a[i]); dp[0][i] = dp[1][i] = 0; } dp[0][0] = dp[1][0] = a[0] = 0; int t = 1; for (int i=1; i<=m; i++) { dp[t][i] = dp[1-t][i-1] + a[i]; int maxn = dp[1-t][i-1]; for (int j=i+1; j<=n-m+i; j++) { maxn = Max(maxn, dp[1-t][j-1]); dp[t][j] = Max(maxn, dp[t][j-1]) + a[j]; } t = 1-t; } t = 1-t; int ans = -1*INF; for (int i=m; i<=n; i++) { if (dp[t][i]>ans) ans = dp[t][i]; } cout << ans << endl; } return 0; }