UVA 11078 Open Credit System（扫描 维护最大值）

Open Credit System

In an open credit system, the students can choose any course they like, but there is a problem. Some of the students are more senior than other students. The professor of such a course has found quite a number of such students who came from senior classes (as if they came to attend the pre requisite course after passing an advanced course). But he wants to do justice to the new students. So, he is going to take a placement test (basically an IQ test) to assess the level of difference among the students. He wants to know the maximum amount of score that a senior student gets more than any junior student. For example, if a senior student gets 80 and a junior student gets 70, then this amount is 10. Be careful that we don't want the absolute value. Help the professor to figure out a solution.

Input
Input consists of a number of test cases T (less than 20). Each case starts with an integer n which is the number of students in the course. This value can be as large as 100,000 and as low as 2. Next n lines contain n integers where the i'th integer is the score of the i'th student. All these integers have absolute values less than 150000. If i < j, then i'th student is senior to the j'th student.

Output
For each test case, output the desired number in a new line. Follow the format shown in sample input-output section.

Sample Input

3
 2
 100
 20
 4
 4
 3
 2
 1

4

1

2

3

4

Output for Sample Input

80
3
-1

题目大意：开放式学分制。给定一个长度为n的整数序列A₀,A₁,...,A_n-1，找出两个整数A_i和A_j（i<j），使得A_i-A_j 尽量大。

分析：使用二重循环会超时。对于每个固定的j，我们应该选择的是小于j且A_i最大的i，而和A_j的具体数值无关。这样，我们从小到大枚举j，顺便维护A_i的最大值即可。

代码如下：

 

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 int A[100000], n;
 5 int main() {
 6   int T;
 7   scanf("%d", &T);
 8   while(T--) {
 9     scanf("%d", &n);
10     for(int i = 0; i < n; i++) scanf("%d", &A[i]);
11     int ans = A[0]-A[1];
12     int MaxAi = A[0]; // MaxAi动态维护A[0]，A[1]，…，A[j-1]的最大值
13     for(int j = 1; j < n; j++) { // j从1而不是0开始枚举，因为j=0时，不存在i
14       ans = max(ans, MaxAi-A[j]);
15       MaxAi = max(A[j], MaxAi); //MaxAi晚于ans更新。想一想，为什么
16     }
17     printf("%d\n", ans);
18   }
19   return 0;
20 }