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.
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.
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».
5 7 1 S.M...M ....... ....... M...M.. ......F
12
7 6 2 S..... ...M.. ...... .....M ...... M..... .....F
11
7 6 2 S..... ...M.. ...... ...... .....M M..... .....F
-1
4 4 2 M... .S.. .... ...F
-1
Please note that monsters can run and kill you on start cell and on finish cell as well.
这道题其实数据还是有点水，用vector居然也能过，下次一定转换为一维做。
题意就是n*m的方格里面能不能到达终点，有意思的是怪兽有攻击方位，先DFS标记处怪兽的攻击范围，然后BFS最短路就解决。
关键点是DFS注意优化不然20组45组数据要超时。
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<map>
#include<string.h>
#include <vector>
#include<queue>
const int maxn=200005;
using namespace std;
vector<char>mat[maxn];
int v[maxn];
int sx,sy,ex,ey;
int m,n,t;
int dirx[]= {-1,1,0,0};
int diry[]= {0,0,1,-1};
struct node
{
int x,y;
} no,ne;
int flag=0;
bool check(int x,int y)
{
if(x>=0&&x<n&&y>=0&&y<m)
return true;
return false;
}
void DFS(int x,int y,int d)
{
if(d==0||flag==1)
return ;
for(int i=0; i<4; i++)
{
int fx=dirx[i]+x;
int fy=diry[i]+y;
if((fx==ex&&fy==ey)||(fx==sx&&fy==sy))
{
flag=1;
return ;//如果起点或者终点在攻击范围直接结束程序
}
if(check(fx,fy))
{
if(mat[fx][fy]=='M')
{
continue;
}
if( v[fx*m+fy]+1<v[x*m+y])//如果即将搜索的点的扩散范围大于这次搜索过来的范围就不管，减少时间复杂度。
{
v[fx*m+fy]=d-1;//如果这个点的当前扩散范围小于即将更新过来的范围就更新。注意每个点不只是搜索一次
DFS(fx,fy,d-1);
}
}
}
}
int BFS()
{
queue<node>q;
no.x=sx,no.y=sy;
q.push(no);
v[sx*m+sy]=0;
while(!q.empty())
{
no=q.front();
q.pop();
for(int i=0; i<4; i++)
{
ne.x=dirx[i]+no.x;
ne.y=diry[i]+no.y;
if(check(ne.x,ne.y))
{
if(v[ne.x*m+ne.y]==-1)
{
v[ne.x*m+ne.y]=v[no.x*m+no.y]+1;
q.push(ne);
}
}
}
}
return v[ex*m+ey];
}
int main()
{
int d;
char s;
scanf("%d %d %d",&n,&m,&d);
t=d;
for(int i=0; i<n; i++)
{
getchar();
for(int j=0; j<m; j++)
{
scanf("%c",&s);
if(s=='S')
{
sx=i,sy=j;
}
if(s=='F')
ex=i,ey=j;
mat[i].push_back(s);
v[i*m+j]=-1;
}
}
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
if(mat[i][j]=='M')
{
if(v[i*m+j]==d)
continue;
v[i*m+j]=d;
DFS(i,j,d);
}
}
if(flag==1)
break;
}
if(flag==1)
cout<<-1<<endl;
else
{
int ans = BFS();
cout<<ans<<endl;
}
return 0;
}