hdu 6019 MG loves gold

最新推荐文章于 2017-10-20 19:04:36 发布

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本文介绍了一道关于挖取不同颜色金子的算法题目。任务目标是最小化使用铲子的数量来挖取所有金子，同时确保不重复挖取相同颜色的金子。文章提供了完整的代码实现，并解释了题目的正确理解方式。

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MG loves gold

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1527 Accepted Submission(s): 629

Problem Description

MG is a lucky boy. He is always be able to find gold from underground.

It is known that the gold is a sequence with

n elements, which has its own color

C .

MG can dig out a continuous area of sequence every time by using one shovel, but he's unwilling to dig the golds of the same color with one shovel.

As a greedy person, he wish to take all the n golds away with least shovel.
The rules also require MG not to dig twice at the same position.

MG thought it very easy and he had himself disdained to take the job. As a bystander, could you please help settle the problem and calculate the answer?

Input

The first line is an integer

T which indicates the case number.(

1<=T<=10 )

And as for each case, there are

1 integer

n in the first line which indicate gold-number(

1<=n<=100000 ).

Then there are

n integers

C in the next line, the x-th integer means the x-th gold’s color(

|C|<=2000000000 ).

Output

As for each case, you need to output a single line.

there should be one integer in the line which represents the least possible number of shovels after taking away all

n golds.

Sample Input

Sample Output

2
3

刚开始题意理解错了，以为求最少可以改变顺序，但真正的意思是按顺序挖金子，遇到相同的换一把铲子，看代码吧

#include<stdio.h>
#include<set>
using namespace std;
int main()
{
    set<int>gold;
    gold.clear();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int c,n,sum=0;
        scanf("%d",&n);
        while(n--)
        {
            scanf("%d",&c);
            if(gold.find(c)!=gold.end())
            {
                sum++;
                gold.clear();
                gold.insert(c);
            }
            else
                gold.insert(c);
        }
        printf("%d\n",sum+1);
        gold.clear();
    }
    return 0;
}