POJ 2479 解题报告

来源：互联网发布：number是什么软件编辑：程序博客网时间：2024/06/03 20:06

这道题是对求一个数组中最大连续子数组和的改编，求两段最大子数组和的和。Leetcode上面好多类似的。看了discuss才回想起来。代码重复的地方很多。其实就是最大子数组和的代码，左右各求一遍。这样对任意位置，都知道左边和右边的最大和是多少，加起来就可以。

2479Accepted720K438MSC++1635B

/* ID: thestor1 LANG: C++ TASK: poj2479 */#include <iostream>#include <fstream>#include <cmath>#include <cstdio>#include <cstring>#include <limits>#include <string>#include <vector>#include <list>#include <set>#include <map>#include <queue>#include <stack>#include <algorithm>#include <cassert>using namespace std;const int MAXN = 50000;int main(){int nums[MAXN];int leftsums[MAXN], rightsums[MAXN];int T;scanf("%d", &T);for (int t = 0; t < T; ++t){int n;scanf("%d", &n);for (int i = 0; i < n; ++i){scanf("%d", &nums[i]);}bool allneg = true;int maxneg = INT_MIN;int sum = 0, maxsum = INT_MIN;for (int i = 0; i < n; ++i){if (allneg && nums[i] < 0){maxneg = max(maxneg, nums[i]);leftsums[i] = maxneg;}else{allneg = false;sum += nums[i];if (sum < 0){sum = 0;}maxsum = max(maxsum, sum);leftsums[i] = maxsum;}}allneg = true;maxneg = INT_MIN;sum = 0, maxsum = INT_MIN;for (int i = n - 1; i >= 0; --i){if (allneg && nums[i] < 0){maxneg = max(maxneg, nums[i]);rightsums[i] = maxneg;}else{allneg = false;sum += nums[i];if (sum < 0){sum = 0;}maxsum = max(maxsum, sum);rightsums[i] = maxsum;}}maxsum = INT_MIN;for (int i = 0; i < n - 1; ++i){if (leftsums[i] + rightsums[i + 1] > maxsum){maxsum = leftsums[i] + rightsums[i + 1];}}printf("%d\n", maxsum);}return 0;  }

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