HDU - 1024 Max Sum Plus Plus(DP)
2017-08-09 10:09
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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.
题解参考http://blog.sina.com.cn/s/blog_677a3eb30100jxqa.html
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <queue> #define N 1000100 #define INF 0x3f3f3f3f using namespace std; long long dp[2] ; long long a ; int main() { int n, m; while(scanf("%d%d", &m, &n) != EOF) { memset(dp, 0, sizeof(dp)); for(int i = 1; i <= n; i++) { scanf("%lld", &a[i]); } for(int i = 1; i <= m; i++) { dp[i % 2][i] = dp[(i + 1)% 2][i- 1] + a[i]; long long int _max = dp[(i + 1) % 2][i - 1]; for(int j = i + 1; j <= n; j++) { _max = max(_max, dp[(i + 1) % 2][j - 1]); dp[i % 2][j] = max(dp[i % 2][j - 1], _max) + a[j]; } } long long int _max = -INF; int index = m % 2; for(int i = m; i <= n; i++) { _max = max(_max, dp[index][i]); } printf("%lld\n", _max); } return 0; }
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