hrbust 1434 Quick out of the Harbour【Bfs+优先队列】

探讨了如何帮助海盗船长迅速带领船只穿越复杂港口的问题。通过BFS算法结合优先队列实现最优路径搜索,考虑了吊桥开关的时间成本。

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Quick out of the Harbour
Time Limit: 3000 MSMemory Limit: 65536 K
Total Submit: 10(6 users)Total Accepted: 8(6 users)Rating: Special Judge: No
Description
Captain Clearbeard decided to go to the harbour for a few days so his crew could inspect and repair the ship. Now, a few days later, the pirates are getting landsick(Pirates get landsick when they don't get enough of the ships' rocking motion. That's why pirates often try to simulate that motion by drinking rum.). Before all of the pirates become too sick to row the boat out of the harbour, captain Clearbeard decided to leave the harbour as quickly as possible.
Unfortunately the harbour isn't just a straight path to open sea. To protect the city from evil pirates, the entrance of the harbour is a kind of maze with drawbridges in it. Every bridge takes some time to open, so it could be faster to take a detour. Your task is to help captain Clearbeard and the fastest way out to open sea.
The pirates will row as fast as one minute per grid cell on the map. The ship can move only horizontally or vertically on the map. Making a 90 degree turn does not take any extra time.
Input
The first line of the input contains a single number: the number of test cases to follow. Each test case has the following format:
1. One line with three integers, h, w (3 <= h;w <= 500), and d (0 <= d <= 50), the height and width of the map and the delay for opening a bridge.
2.h lines with w characters: the description of the map. The map is described using the following characters:
—"S", the starting position of the ship.
—".", water.
—"#", land.
—"@", a drawbridge.
Each harbour is completely surrounded with land, with exception of the single entrance.
Output
For every test case in the input, the output should contain one integer on a single line: the travelling time of the fastest route to open sea. There is always a route to open sea. Note that the open sea is not shown on the map, so you need to move outside of the map to reach open sea.
Sample Input
26 5 7######S..##@#.##...##@####.###4 5 3######S#.##@..####@#
Sample Output
1611
Source
BAPC 2011

题目大意:

S表示船的起点,@代表可以打开的吊桥,不过想要打开吊桥需要时间d,问你船想离开这个地图的最小花费时间。


思路:


1、因为有吊桥的出现,我们Bfs优先从时间+最短路径优先,变成了最短路径优先,然而我们显然需要的是时间优先,所以加入一个优先队列元素,使得队头时间最小,按照时间优先的顺序来Bfs即可。


2、注意地图大小,注意边界判定。


Ac代码:

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
struct node
{
    int x,y,time;
    friend bool operator <(node a,node b)
    {
        return a.time>b.time;
    }
}now,nex;
char a[651][651];
int vis[651][651];
int fx[4]={0,0,1,-1};
int fy[4]={1,-1,0,0};
int n,m,k;
void Bfs_priority(int x,int y)
{

    memset(vis,0,sizeof(vis));
    priority_queue<node>s;
    vis[x][y]=1;
    now.x=x;
    now.y=y;
    now.time=0;
    s.push(now);
    while(!s.empty())
    {
        now=s.top();
        if(now.x==0||now.x==n-1||now.y==0||now.y==m-1)
        {
            printf("%d\n",now.time+1);
            return ;
        }
        s.pop();
        for(int i=0;i<4;i++)
        {
            nex.x=now.x+fx[i];
            nex.y=now.y+fy[i];
            if(nex.x>=0&&nex.x<n&&nex.y>=0&&nex.y<m&&a[nex.x][nex.y]!='#'&&vis[nex.x][nex.y]==0)
            {
                if(a[nex.x][nex.y]=='@')
                {
                    nex.time=now.time+k+1;
                }
                else nex.time=now.time+1;
                vis[nex.x][nex.y]=1;
                s.push(nex);
            }
        }
    }
    return ;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int sx,sy;
        scanf("%d%d%d",&n,&m,&k);
        for(int i=0;i<n;i++)
        {
            scanf("%s",a[i]);
            for(int j=0;j<m;j++)
            {
                if(a[i][j]=='S')
                {
                    sx=i;
                    sy=j;
                }
            }
        }
        Bfs_priority(sx,sy);
    }
}
/*
#####
#...#
#@#.#
#...#
#@###
#S###
*/




资源下载链接为: https://pan.quark.cn/s/67c535f75d4c 在机器人技术中,轨迹规划是实现机器人从一个位置平稳高效移动到另一个位置的核心环节。本资源提供了一套基于 MATLAB 的机器人轨迹规划程序,涵盖了关节空间和笛卡尔空间两种规划方式。MATLAB 是一种强大的数值计算与可视化工具,凭借其灵活易用的特点,常被用于机器人控制算法的开发与仿真。 关节空间轨迹规划主要关注机器人各关节角度的变化,生成从初始配置到目标配置的连续路径。其关键知识点包括: 关节变量:指机器人各关节的旋转角度或伸缩长度。 运动学逆解:通过数学方法从末端执行器的目标位置反推关节变量。 路径平滑:确保关节变量轨迹连续且无抖动,常用方法有 S 型曲线拟合、多项式插值等。 速度和加速度限制:考虑关节的实际物理限制,确保轨迹在允许的动态范围内。 碰撞避免:在规划过程中避免关节与其他物体发生碰撞。 笛卡尔空间轨迹规划直接处理机器人末端执行器在工作空间中的位置和姿态变化,涉及以下内容: 工作空间:机器人可到达的所有三维空间点的集合。 路径规划:在工作空间中找到一条从起点到终点的无碰撞路径。 障碍物表示:采用二维或三维网格、Voronoi 图、Octree 等数据结构表示工作空间中的障碍物。 轨迹生成:通过样条曲线、直线插值等方法生成平滑路径。 实时更新:在规划过程中实时检测并避开新出现的障碍物。 在 MATLAB 中实现上述规划方法,可以借助其内置函数和工具箱: 优化工具箱:用于解决运动学逆解和路径规划中的优化问题。 Simulink:可视化建模环境,适合构建和仿真复杂的控制系统。 ODE 求解器:如 ode45,用于求解机器人动力学方程和轨迹执行过程中的运动学问题。 在实际应用中,通常会结合关节空间和笛卡尔空间的规划方法。先在关节空间生成平滑轨迹,再通过运动学正解将关节轨迹转换为笛卡
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