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最新推荐文章于 2024-04-24 23:16:35 发布

climber_studio

最新推荐文章于 2024-04-24 23:16:35 发布

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文章标签：算法优化

本文链接：https://blog.youkuaiyun.com/climber_studio/article/details/5813899

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最大连续和问题，有多种的复杂度的算法O(N^3)、O(N^2)、O(NlogN)、O(n).

下面从这里引用关于上文算法的介绍，增加了一些内容：

1.最原始，最古老，最暴力的方法，时间复杂度是T(n) = O(n^3)

best = a[1]; for(i = 1; i <= n; i++) for(j = i; j <= n; j++) { int sum = 0; for(k = i; k <= j; k++) sum += a[k]; if(sum > best) best = sum; }
2.第一次优化：利用“连续子序列之和等于两个前缀和之差”，时间复杂度为：T(n) = O(n^2)：
best = s[1]; for(int i = 1; i <= n; i++) for(int j = i; j<= n; j++) best = (best > (s[j] - s[i])? best : (s[j] - s[i])); //更新最大值
3.再次优化：分治法，时间复杂度：

int maxsum(int *a, int x, int y) { int i, m, v, L, R, max; if(y-x == 1) return a[x]; m = x + (y-x)/2; max = maxsum(a, x, m) > maxsum(a, m, y) ? maxsum(a, x, m) : maxsum(a, m, y); v = 0; L = a[m-1]; for(int i = m-1; i >= x; i--) { v += a[i]; L = L > v? L: v; } v = 0; R = a[m]; for(int i = m; i < y; i++) { v += a[i]; R = R > v? R: v; } return max > (L+R)? max: (L+R); }
4线性DP,时间复杂度:T(n) = O(n),方程sum[i]=max{sum[i-1]+a[i],a[i]}:

sum = a[1], temp = a[1]; for(int i = 2; i <= n; i++) { if(sum > 0) temp += a[i]; else temp = a[i]; if(temp > sum) sum = temp; }

接下来回到题目，题目要求的子序列尽量长，尽量靠前，且输出格式很恶心.-.-纠结2h都没AC，之后参考F.u的程序终于AC了...程序30+行（我在优快云看到某位搞ACM的程序近80行- -||）

 
  #include<iostream>
using namespace std;
int s[20010] = {0};
int main()
{
    int N, n, i, j, k, S, E, len, ans, tmp;
    cin >> N;
    for (j = 0; j < N; j++)
    {
        cin >> n;
        for (i = 1; i < n; i++) cin >> s[i];
        k = 1; tmp = ans = len = 0;
        for(i = 1; i < n; i++)
        {
            if(tmp + s[i] < 0)
            {
                tmp = 0;
                k = i+1;
            }
            else
            {
                tmp += s[i];
                if (tmp > ans || (tmp == ans && i-k+1 > len))
                {
                    S = k; E = i;
                    ans = tmp; len= i-k+1;
                }
            }
        }
        E++;
        if (ans == 0 ) printf("Route %d has no nice parts/n", j+1);
        else printf("The nicest part of route %d is between stops %d and %d/n", j+1, S, E);
    }
} 
 
 