Gym - 101755H

题目
You play a new RPG. The world map in it is represented by a grid of n × m cells. Any playing character staying in some cell can move from this cell in four directions — to the cells to the left, right, forward and back, but not leaving the world map.

Monsters live in some cells. If at some moment of time you are in the cell which is reachable by some monster in d steps or less, he immediately runs to you and kills you.

You have to get alive from one cell of game field to another. Determine whether it is possible and if yes, find the minimal number of steps required to do it.

Input
The first line contains three non-negative integers n, m and d (2 ≤ n·m ≤ 200000, 0 ≤ d ≤ 200000) — the size of the map and the maximal distance at which monsters are dangerous.

Each of the next n lines contains m characters. These characters can be equal to «.», «M», «S» and «F», which denote empty cell, cell with monster, start cell and finish cell, correspondingly. Start and finish cells are empty and are presented in the input exactly once.

Output
If it is possible to get alive from start cell to finish cell, output minimal number of steps required to do it. Otherwise, output «-1».
题目大意
有一个NM的矩阵,每个格子可能为’.’,‘M’,‘S’,'F’中的一种。'M’为敌人,其攻击范围为距离它小于等于d的所有格子,‘S’为起点,'F’为终点,‘.’为空格子。求一个人从’S’到‘F’的最短路径。
解题思路
由于数据范围(n
m<=200000),所以可以用一维数组在表示每个格子
例如点(i,j)可以表示为a[i*m+j]
先将敌人M的攻击范围用BFS标记,再用BFS搜索从’S’到’F’的最短路径。
具体代码

#include <cstdio>
#include <vector>
#include <queue>
#include <iostream>
using namespace std;
queue <int> X,Y,XX,YY;
int r[404040];
bool sym[404040];
int d[404040];
int n,m,L,sx,sy,ex,ey;
int ans=-1;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
bool symt=false;
int S=1;
void bfs1()
{
	while (!XX.empty())
	{
	    int x=XX.front();
		int y=YY.front();
		XX.pop();YY.pop();
		for (int i=0;i<4;i++)
		{
			int xx=x+dx[i];
			int yy=y+dy[i];
			if (xx<1 || xx>n || yy<1 || yy>m || d[x*m+y]==0) continue;	
			if (d[x*m+y]-1>=d[xx*m+yy])
			{
				d[xx*m+yy]=d[x*m+y]-1;
				if (!sym[xx*m+yy])
				{
					sym[xx*m+yy]=true;
					XX.push(xx);
					YY.push(yy);
				}
			}
		}
	}
}
void BFS()
{
	X.push(sx);
	Y.push(sy);
	sym[sx*m+sy]=1;
	while (!X.empty())
	{
		int x=X.front();
		int y=Y.front();
		X.pop();Y.pop();
		for (int i=0;i<4;i++)
		{
			int xx=x+dx[i];
			int yy=y+dy[i];
			if (xx<1 || xx>n || yy<1 || yy>m || sym[xx*m+yy]!=0) continue;
			X.push(xx);
			Y.push(yy);
			sym[xx*m+yy]=1;
			r[xx*m+yy]=r[x*m+y]+1;
			if (xx==ex && yy==ey) 
			{
				ans=r[xx*m+yy];
				return;
			}
		}
	}
}
int main()
{
	scanf("%d%d%d",&n,&m,&L);
    for (int i=1;i<=n;i++)
      for (int j=1;j<=m;j++)
        {
        	char ch;
        	cin>>ch;
            if (ch=='S')
        	{
        		sx=i;
        		sy=j;
        	
        	}
        	if (ch=='F')
        	{
        		ex=i;
        		ey=j;
        	}
        	if (ch=='M') 
        	{
        		XX.push(i);
        		YY.push(j);
        		d[i*m+j]=L;
        		sym[i*m+j]=true;
        	}
        } 
      bfs1();	  
      if (sym[sx*m+sy]!=0 || sym[ex*m+ey]!=0) ans=-1;     
	  else BFS();  
	  printf("%d\n",ans);
} 
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转载自blog.csdn.net/weixin_45723759/article/details/103953416