[zoj] 4178. Killing the Brute-force

题目

Chenjb is the task author of the 71-st Zhejiang Provincial Collegiate Programming Contest. He created an NP-Hard problem on a graph with n vertices, the intended solution of which is DFS with branching and pruning. Chenjb is an experienced task author, he knows that the constraints can't be too tight, so he will set the time limit t to be 3x, where x is the running time of the intended solution on the slowest test case.

Chenjb heard that the contest would be unfortunately held on Potato OJ, whose hardware is not so good. To protect Potato OJ from being overloaded and the contest being a failure, Chenjb plans to cut down the input size of his task. He has tested the running time of the intended solution and the brute-force solution for various values of n. Note that for simplicity, we assume the running time of the solutions on a test case only depends on the value of n.

Please help Chenjb find the smallest value of n such that 1≤n≤m, and the brute-force solution won't pass the problem. Note that if the time limit is t, we assume that a solution will pass if and only if its running time on each test case is smaller than or equal to t.

Input:

The input contains multiple cases. The first line of the input contains a single integer T (1≤T≤50), the number of cases.

For each case, the first line of the input contains a single integer m (1≤m≤50), denoting the upper bound of n.

The second line contains m integers a1,a2,…,a**m (1≤a**i≤100,a**i≤a**i+1), the i-th of which denotes the running time of the intended solution when n=i.

The third line contains m integers b1,b2,…,b**m (1≤b**i≤100,b**i≤b**i+1), the i-th of which denotes the running time of the brute-force solution when n=i.

Output:

For each case, print a single line containing a single integer denoting the minimum possible value of n. If there is no solution, print −1 instead.

Sample Input:

Sample Output:

3

-1

思路

该题目大意是在说Chenjb老师在为土豆OJ设计问题通过门槛，由于OJ硬件配置太低，没有办法运行太多用例。现需要裁剪用例个数，要求对每个问题求出过滤掉暴力算法时所需要的最少用例数量。

这个问题乍一看摸不着头脑，实际上却很简单。。。针对一道题目，从左到右遍历用例，找到第一个时间限制（3*理想用例耗时）小于暴力算法耗时的用例为止，已遍历用例的个数就是所需要的值。以示例中的第一道题目为例，从左到右遍历用例，遍历至第3个用例时，发现时间限制（3*7=21）小于暴力算法耗时（30），那么这道用例就能够把暴力算法过滤掉，因此最终结果为3。

代码

python代码：

T=input()

T=int(T)

for i in range(T):

    m=input()

    intend=input().split()

    brute=input().split()

    intend=[int(i) for i in intend]

    brute=[int(i) for i in brute]

    n=-1

    for i in range(len(intend)):

        if intend[i]*3 < brute[i]:

            n=i+1

            break

    print(n)