War Chess bfs+优先队列

War chess is hh's favorite game:
In this game, there is an N * M battle map, and every player has
his own Moving Val (MV). In each round, every player can move in four
directions as long as he has enough MV. To simplify the problem, you are
given your position and asked to output which grids you can arrive.




In the map:

'Y' is your current position (there is one and only one Y in the given map).

'.' is a normal grid. It costs you 1 MV to enter in this gird.

'T' is a tree. It costs you 2 MV to enter in this gird.

'R' is a river. It costs you 3 MV to enter in this gird.

'#' is an obstacle. You can never enter in this gird.

'E's are your enemies. You cannot move across your enemy, because
once you enter the grids which are adjacent with 'E', you will lose all
your MV. Here “adjacent” means two grids share a common edge.

'P's are your partners. You can move across your partner, but you
cannot stay in the same grid with him final, because there can only be
one person in one grid.You can assume the Ps must stand on '.' . so ,it
also costs you 1 MV to enter this grid.

InputThe first line of the inputs is T, which stands for the number of test cases you need to solve.

Then T cases follow:

Each test case starts with a line contains three numbers N,M and MV
(2<= N , M <=100,0<=MV<= 65536) which indicate the size of
the map and Y's MV.Then a N*M two-dimensional array follows, which
describe the whole map.OutputOutput the N*M map, using '*'s to replace all the grids 'Y'
can arrive (except the 'Y' grid itself). Output a blank line after each
case.Sample Input

5
3 3 100
...
.E.
..Y

5 6 4
......
....PR
..E.PY
...ETT
....TT

2 2 100
.E
EY

5 5 2
.....
..P..
.PYP.
..P..
.....

3 3 1
.E.
EYE
...

Sample Output

...
.E*
.*Y

...***
..**P*
..E*PY
...E**
....T*

.E
EY

..*..
.*P*.
*PYP*
.*P*.
..*..

.E.
EYE
.*.

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
int dir[][]={,,,,,-,-,};
int t,n,m,val,head,tail,tx,ty,d;
char map[][];

int vis[][];
struct que
{
    int x,y,mv;
    friend bool operator <(que a,que b)
    {
        return a.mv<b.mv;
    }
}cur;
int judge(int x,int y)
{
    if(x<||y<||x>=n||y>=m)return ;
    return ;
}
int check(int x,int y)
{
    for(int i=;i<;i++)
    if(judge(x+dir[i][],y+dir[i][])&&map[x+dir[i][]][y+dir[i][]]=='E')return ;
    return ;
}
int check1(int x,int y)
{
    if(vis[x][y]<)return ;
    if(map[x][y]=='P'||map[x][y]=='Y')return ;
    return ;
}
int main()
{
    priority_queue <que>q;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&val);
        head=tail=;
        memset(vis,-,sizeof(vis));
        for(int i=;i<n;i++)
        scanf("%s",map[i]);
        for(int i=;i<n;i++)
        {
            for(int j=;j<m;j++)
            {
                if(map[i][j]=='Y'){
                cur.x=i,cur.y=j;
                cur.mv=val;
                q.push(cur);
                vis[i][j]=val;
                break;}
            }
        }

        while(!q.empty())
        {
            for(int i=;i<;i++)
            {
                tx=q.top().x+dir[i][];
                ty=q.top().y+dir[i][];
                if(!judge(tx,ty)||map[tx][ty]=='#'||map[tx][ty]=='E')continue;
                if(map[tx][ty]=='T')d=q.top().mv-;
                else if(map[tx][ty]=='R')d=q.top().mv-;
                else d=q.top().mv-;
                if(check(tx,ty)&&d>)d=;
                if(d>vis[tx][ty])
                {
                    vis[tx][ty]=d;
                    if(d>)
                    {cur.x=tx;
                    cur.y=ty;
                    cur.mv=d;
                    q.push(cur);}
                }
            }
            q.pop();
        }

        for(int i=;i<n;i++)
        {
            for(int j=;j<m;j++)
            {
                if(check1(i,j))putchar('*');
                else putchar(map[i][j]);
            }
            cout<<endl;
        }
        cout<<endl;
    }
}