UCF Local Programming Contest 2017 D. Editor Navigation题解

题目链接https://nanti.jisuanke.com/t/44820

题目太长就不放了，大概意思是，给你光标的现在位置和目标位置，问你最短多少步能到达目标位置。

运用算法：bfs
注意：我觉得bfs重点就在于判断条件这里一定要思路清晰。

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具体看代码，代码有注释。

AC代码：

#include<cstdio>
#include<string>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
#include<ctime>
#include<queue>

using namespace std;

int num[130];//记录每行的长度
int vis[130][90];//标记是否走过此格 
int dis[10][2] = {-1,0,0,-1,0,1,1,0};//上下左右四个点 
int res = -1;
int nx,ny;//目标点 
int n;

struct node{
    int x,y;
    int cnt = 0;
}f[10];

bool check(int &x,int &y)
{
    if(x < 1||x > n||vis[x][y] == 1)return true;
    else return false;
}

queue<node>q;
int bfs()
{
    node pre;
    while(!q.empty())
    {
        pre = q.front();
        q.pop();
        node now;
        for(int i = 0;i < 4;i++)
        {
            now.x = pre.x+dis[i][0];
            now.y = pre.y+dis[i][1];
            if(dis[i][0]!=0)		//上下走 
            {
                if(now.x >= 1&&now.x <= n&&now.y > num[now.x])
                {
                    now.y = num[now.x];
                }
            }
            else if(dis[i][1]!=0)	//左右走 
            {
                if(now.y == num[now.x]+1&&now.x >= 1&&now.x <= n)
                {
                     now.x++;
                     if(now.x >= 1&&now.x <= n)now.y = 0;
                }
                else if(now.y == -1&&now.x >= 1&&now.x <= n)
                {
                    now.x--;
                    if(now.x >= 1&&now.x <= n)now.y = num[now.x];
                }
            }   
      
            if(check(now.x,now.y))
            {
                continue;
            }
   
            vis[now.x][now.y] = 1;
            now.cnt = pre.cnt+1;
            if (now.x == nx&&now.y == ny)//判断是否找到目标点 
            {
                res = now.cnt;
                break;//根据队列先进先出的性质，找到的第一个点就是最步数 
            }
            q.push(now);
        }
        if(res!=-1)break;
    }
}

int main()
{
    int t;
    scanf/("%d",&t);
    while(t--)
    {
         res = -1;
         memset(vis,0,sizeof(vis));	//初始化标记数组
         scanf("%d",&n);
         for(int i = 1 ; i <= n;i++)
         {
   	     scanf("%d",&num[i]);
  	 }
         scanf("%d %d",&f[0].x,&f[0].y);
         scanf("%d %d",&nx,&ny);
         //以上是输入
         vis[f[0].x][f[0].y] = 1;
  	 q.push(f[0]);
  	 if(f[0].x == nx&&f[0].y == ny)//如果初始点即为目标点时
  	 {
  	     printf("0\n");
         }
  	 else 
  	 {
   	     bfs();
   	     printf("%d\n",res);
 	 }
    	 while(!q.empty())//清空队列 
 	 {
  	     q.pop();
 	 }
    }
    return 0;
}