11573 - Ocean Currents DFS变形

每次做这种题总是差一点，还是能力不够吧。。。需要用优先队列，这点我想到了，主要是没有想到利用一个vis数组，更新每个点的最短距离，贪心思想有问题。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<queue>
#define MAXD 1000 + 10
using namespace std;
struct Point{
    int x;
    int y;
    int d;
    Point(int a,int b,int c){
        x = a; y = b ; d = c;
    }
    friend bool operator < (Point p,Point q){
        if(p.d > q.d) return true;
        else
        return false;
    }
};
int  dir[][2] ={{-1,0},{-1,1},{0,1},
                {1,1},{1,0},{1,-1},
                {0,-1},{-1,-1}};
int  n,m;
int  G[MAXD][MAXD];
int  vis[MAXD][MAXD];
int  dist[MAXD][MAXD];
int  sx,sy,ex,ey;
void solve(){
    memset(vis,-1,sizeof(vis));
    priority_queue<Point>q;
    q.push(Point(sx,sy,0));
    vis[sx][sy] = 0;
    while(!q.empty()){
        int x = q.top().x , y = q.top().y;
        int d = q.top().d;
        if(x == ex && y == ey){
            printf("%d\n",d);
            return ;
        }
        q.pop();
        int t = G[x][y];
        for(int i = 0 ; i < 8; i++){
            int nx = x + dir[i][0];
            int ny = y + dir[i][1];
            if(nx >=0 && nx < n && ny >= 0 && ny < m){
                int _d;
                if(i == t)
                _d = d;
                else
                _d = d + 1;
                if(vis[nx][ny] == -1 || _d < vis[nx][ny]){
                    vis[nx][ny] = _d;
                    q.push(Point(nx,ny,_d));
                }
            }
        }
    }
}
int main(){
    while(scanf("%d%d",&n,&m) != EOF){
    for(int i = 0 ; i < n ; i++){
        char str[MAXD];
        scanf("%s",str);
        for(int j = 0 ; j < m ; j++)
        G[i][j] = str[j] - '0';
    }
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d%d",&sx,&sy,&ex,&ey);
        sx -- ; sy -- ; ex -- ; ey--;
        solve();
    }
    }
    return 0;
}