codeforces1064D

题意是给你一个字符矩阵，‘*’是障碍物，给你起点r,c，和最多能向左走的步数x,最多能向右走的步数y

问你可访问的点的个数是多少

一开始用直接暴力BFS被hack了，后来加上优先队列就能ac

#include<bits/stdc++.h>
#define ll long long
using namespace std;
char data[2005][2005];
int n,m,r,c,x,y;
struct node
{
    int x,y,l,r;
    bool operator <(node a)const
    {
        if(l==a.l)
            return r>a.r;
        else
            return l>a.l;
    }
};
int ans=0;
bool isok(int x,int y)
{
    return x>=0&&x<n&&y>=0&&y<m&&data[x][y]!='*';
}
int vis[2005][2005];
int dir[4][2]= {-1,0,0,1,1,0,0,-1};
void bfs()
{
    priority_queue<node>q;
    q.push(node{r-1,c-1,0,0});
    vis[r-1][c-1]=1;
//    data[r-1][c-1]='+';
    ans++;
    while(!q.empty())
    {
        node u = q.top();
        q.pop();
//        printf("%d %d
",u.x,u.y);
        for (int i=0; i<4 ; i++ )
        {
            int newx=u.x+dir[i][0];
            int newy=u.y+dir[i][1];
            if(isok(newx,newy) && !vis[newx][newy] && (u.l+(i==3?1:0)<=x) && (u.r+(i==1?1:0)<=y) )
            {
                ans++;
//                data[newx][newy]='+';
                vis[newx][newy]=1;
                q.push(node{newx,newy,u.l+(i==3?1:0),u.r+(i==1?1:0)});
            }
        }
    }
}
int main()
{
    ans=0;
    memset(vis,0,sizeof(vis));
    scanf("%d%d",&n,&m);
    scanf("%d%d",&r,&c);
    scanf("%d%d",&x,&y);
    getchar();
    for (int i=0; i<n ; i++ )
    {
        for (int j=0; j<m ; j++ )
        {
            data[i][j]=getchar();
        }
        getchar();
    }
    bfs();
    printf("%d",ans);
}