Day 5 E. Arranging The Sheep

这是一个关于贪心算法的游戏问题,名为'Arranging The Sheep'。玩家需要通过最少的移动次数使羊排列整齐,无空格间隔。对于给定的羊的位置字符串,求解达到排列整齐的最小移动次数。示例和中位数定理被用来解释解决问题的策略,即找到中间的*作为基准进行移动。

Problem:
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:

the sheep at the 4 position moves to the right, the state of the level: " * * . .* . ";
**the sheep at the 2 position moves to the right, the state of the level: " * . * . * . " ;
the sheep at the 1 position moves to the right, the state of the level: " . * * . * . ";
the sheep at the 3 position moves to the right, the state of the level: " . * . * * . ";
the sheep at the 2 position moves to the right, the state of the level: " . . * * * . ";
the sheep are lined up and the game ends.
For a given level, determine the minimum number of moves you need to make to complete the level.

Input
The first line contains one integer t (1≤t≤10^4). Then t test cases follow.
The first line of each test case contains one integer n (1≤n≤10^6).
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 n over all test cases does not exceed 10^6.

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

Example
input
5

6
* * . * . .

5
* * * * *

3
. * .

3
. . .

10
* . * . . . * . * *

output
1

0

0

0

9

题目大致意思为:给你一串长为n 仅由 *和.组成的字符串 求至少经过多少次变换 *号能排成一行 每次变换只能左右移动一个 *的位置 这里我用到了中位数定理…
中位数定理
背景:如果有一个数轴 数轴上有若干个点 现要在数轴上找一点 使得它到各个点的距离之和最短。
结论: 中位数就是最优解
中位数有这样的性质: 所有数与中位数的绝对差之和最小
中位数是数列中间的那个数 或者是中间的那两个数之一.
在这道题目中 中位数就是中间 *的位数

代码如下

#include <iostream>
#
C - Giant Domino Editorial  /   Time Limit: 2 sec / Memory Limit: 1024 MiB Score : 300 points Problem Statement There are 𝑁 dominoes numbered from 1 to 𝑁 . The size of domino 𝑖 is 𝑆 𝑖 . Consider arranging some dominoes in a line from left to right and then toppling them. When domino 𝑖 falls to the right, if the size of the domino placed immediately to the right of domino 𝑖 is at most 2 𝑆 𝑖 , then that domino also falls to the right. You decided to choose two or more dominoes and arrange them in a line from left to right. The arrangement of dominoes must satisfy the following conditions: • The leftmost domino is domino 1 .The rightmost domino is domino 𝑁 . • When only domino 1 is toppled to the right, domino 𝑁 eventually falls to the right as well. Does an arrangement of dominoes satisfying the conditions exist? If it exists, what is the minimum number of dominoes that need to be arranged? You are given 𝑇 test cases, solve the problem for each of them. Constraints • 1 ≤ 𝑇 ≤ 1 0 5 • 2 ≤ 𝑁 ≤ 2 × 1 0 5 • 1 ≤ 𝑆 𝑖 ≤ 1 0 9 • The sum of 𝑁 over all test cases is at most 2 × 1 0 5 . • All input values are integers.  Input The input is given from Standard Input in the following format, where c a s e 𝑖 means the 𝑖 -th test case: 𝑇 c a s e 1 c a s e 2 ⋮  c a s e 𝑇 Each test case is given in the following format: 𝑁 𝑆 1 𝑆 2 … 𝑆 𝑁 Output Output 𝑇 lines. The 𝑖 -th line should contain the answer for the 𝑖 -th test case. For each test case, if there is no arrangement of dominoes satisfying the conditions, output -1; otherwise, output the minimum number of dominoes to arrange.  Sample Input 1Copy Copy 3 4 1 3 2 5 2 1 100 10 298077099 766294630 440423914 59187620 725560241 585990757 965580536 623321126 550925214 917827435 Sample Output 1Copy Copy 4 -1 3 For the 1 st test case, arranging the dominoes from left to right in the order domino 1 , domino 3 , domino 2 , domino 4 satisfies the conditions in the pr
最新发布
11-20
以下是使用 C 语言解决该多米诺骨牌排列问题的代码: ```c #include <stdio.h> #include <stdlib.h> #define MAX_N 1005 #define INF 0x3f3f3f3f int n; int graph[MAX_N][MAX_N]; int dist[MAX_N]; int visited[MAX_N]; // 迪杰斯特拉算法求最短路径 void dijkstra() { for (int i = 1; i <= n; i++) { dist[i] = INF; visited[i] = 0; } dist[1] = 1; for (int i = 1; i <= n; i++) { int min_dist = INF, u = -1; for (int j = 1; j <= n; j++) { if (!visited[j] && dist[j] < min_dist) { min_dist = dist[j]; u = j; } } if (u == -1) break; visited[u] = 1; for (int v = 1; v <= n; v++) { if (!visited[v] && graph[u][v] && dist[u] + 1 < dist[v]) { dist[v] = dist[u] + 1; } } } } int main() { while (scanf("%d", &n) != EOF) { // 初始化图 for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { graph[i][j] = 0; } } // 读入骨牌推倒关系 for (int i = 1; i <= n; i++) { int k; scanf("%d", &k); for (int j = 0; j < k; j++) { int v; scanf("%d", &v); graph[i][v] = 1; } } // 计算最短路径 dijkstra(); // 输出结果 if (dist[n] == INF) { printf("0\n"); } else { printf("%d\n", dist[n]); } } return 0; } ``` 上述代码使用迪杰斯特拉算法来解决多米诺骨牌排列问题。首先,将图的邻接矩阵 `graph` 初始化为 0,表示没有边相连。然后,读入每个骨牌可以推倒的其他骨牌信息,更新图的邻接矩阵。接着,使用迪杰斯特拉算法计算从 1 号骨牌到 N 号骨牌的最短路径,最短路径的长度即为满足条件的最少骨牌数量。如果无法从 1 号骨牌推倒 N 号骨牌,则输出 0。
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