Arranging The Sheep-优快云博客

本文介绍了游戏'安排羊群'的玩法，目标是通过最少的移动次数使羊群整齐排列。文章给出了一个示例，展示了如何在特定关卡中通过策略完成任务，并提供了一个C++程序来计算完成关卡所需的最小移动次数。程序读取关卡描述，计算并输出结果。

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You are playing the game “Arranging The Sheep”. The goal of this game is to make the sheep line up. The level in the game is described by a string of length n, consisting of the characters ‘.’ (empty space) and ‘*’ (sheep). In one move, you can move any sheep one square to the left or one square to the right, if the corresponding square exists and is empty. The game ends as soon as the sheep are lined up, that is, there should be no empty cells between any sheep.

For example, if n=6 and the level is described by the string “**.*…”, then the following game scenario is possible:

1.the sheep at the 4 position moves to the right, the state of the level: “**…*.”;

2.the sheep at the 2 position moves to the right, the state of the level: “..*.”;

3.the sheep at the 1 position moves to the right, the state of the level: “.**.*.”;

4.the sheep at the 3 position moves to the right, the state of the level: “.*.**.”;

5.the sheep at the 2 position moves to the right, the state of the level: “…***.”;

6.the sheep are lined up and the game ends.

输入

The first line contains one integer t(1≤t≤104). Then tt test cases follow.

The first line of each test case contains one integer n(1≤n≤106).

The second line of each test case contains a string of length n, consisting of the characters ‘.’ (empty space) and ‘*’ (sheep) — the description of the level.

It is guaranteed that the sum of nn over all test cases does not exceed 106.

输出

For each test case output the minimum number of moves you need to make to complete the level.

样例输入

5
6
**.*…
5

3
..
3
…
10
.….**

样例输出

1
0
0
0
9

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+5;
int n;
int a[N],b[N];
char s[N];
int main(){
	cin>>n;
    while(n--)
    {
    	int k=0;
    	int t;
    	cin>>t;
       scanf("%s",s);
     for(int i=0;i<t;i++)
     {
     	if(s[i]=='*')
     	{
     		a[k]=i+1;
			 b[k]=k; 
			 k++;
		 }
	 }
	 ll res=0;
	 ll sum=0;
	 for(int i = 0; i<k; i ++) 
	 {
	 res += abs(a[i]-a[k/2]);
	 sum+=abs(b[i]-b[k/2]);
     }  
    printf("%lld\n",res-sum);
}
    return 0;
}