灵能传输(前缀和 + 贪心)

1248. 灵能传输

灵能传输

思路

1.观察发现每次对一个圣堂武士\(a_{i}\)一次灵能传输都是对前缀和\(S_{i - 1}\),\(S_{i}\)得一次交换
2.问题转换为\(max(S_{i} - S_{i - 1})\)的最小值
3.贪心策略可知只有当\(S\)单调时才能使得\(max(S_{i} - S_{i - 1})\)最小
4.由于题目条件使得\(S_{0}\) 和 \(S_{n}\) 得位置固定不变，但是\(S_{0}\) 和 \(S_{n}\)不一定是最大值或者最小值，所以贪心策略发生改变

样例输入：

3
4
-1 -2 -3 7
4
2 3 4 -8
5
-1 -1 6 -1 -1

样例输出：

5
7
4

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 300010;

int n;
LL a[N], s[N];
bool st[N];

int main()
{
    int T;
    scanf("%d", &T);
    while (T -- )
    {
        scanf("%d", &n);
        s[0] = 0;
        for (int i = 1; i <= n; i ++ )
        {
            scanf("%lld", &a[i]);
            s[i] = s[i - 1] + a[i];
        }

        LL s0 = s[0], sn = s[n];
        if (s0 > sn) swap(s0, sn);
        sort(s, s + n + 1);

        for (int i = 0; i <= n; i ++ )
            if (s[i] == s0)
            {
                s0 = i;
                break;
            }

        for (int i = n; i >= 0; i -- )
            if (s[i] == sn)
            {
                sn = i;
                break;
            }

        memset(st, 0, sizeof st);
        int l = 0, r = n;
        for (int i = s0; i >= 0; i -= 2)
        {
            a[l ++ ] = s[i];
            st[i] = true;
        }
        for (int i = sn; i <= n; i += 2)
        {
            a[r -- ] = s[i];
            st[i] = true;
        }
        for (int i = 0; i <= n; i ++ )
            if (!st[i])
                a[l ++ ] = s[i];

        LL res = 0;
        for (int i = 1; i <= n; i ++ ) res = max(res, abs(a[i] - a[i - 1]));

        printf("%lld\n", res);
    }

    return 0;
}

---