电路布线问题-分支限界方法。到底哪里错了啊

Question

#include<iostream>
#include<queue>
using namespace std;
int m,n,ax,ay,bx,by;
int g[500][500];
int len;
struct node{
    int x;
    int y;
};
void wire(){  //分支限界，只要找到可行解必是最优解
    if(ax==bx && ay==by){
        len=0;
    }
    queue<node> q;
    node offset[4];
    offset[0].x=0; offset[0].y=1;  //right
    offset[1].x=1; offset[1].y=0;  //down
    offset[2].x=0; offset[2].y=-1;  //left
    offset[3].x=-1; offset[3].y=1;  //up
    int neighbour_num=4;
    node here,neighbour;
    here.x=ax; here.y=ay;
    g[ax][ay]=2;
    do{
        for(int i=0;i<neighbour_num;i++){
            neighbour.x=here.x+offset[i].x;
            neighbour.y=here.y+offset[i].y;
            if(g[neighbour.x][neighbour.y]==0){  //该方格未被标记
                g[neighbour.x][neighbour.y]=g[here.x][here.y]+1;
                if((neighbour.x == bx) && (neighbour.y == by)){
                    //布线完成
                    break;
                }
                q.push(neighbour);
            }
        }
        if((neighbour.x == bx) && (neighbour.y == by)){
            //布线完成
            break;
        }
        if(q.empty()){
            break;
        }
        here=q.front();
        q.pop();
    }while(1);
    return ;
}
int main(){
    cin>>m>>n;
    for(int i=0;i<=n;i++){
        g[0][i]=1;
        g[m+1][i]=1;
    }
    for(int i=0;i<=m;i++){
        g[i][0]=1;
        g[i][n+1]=1;
    }
    for(int i=1;i<=m;i++){
        for(int j=1;j<=n;j++){
            cin>>g[i][j];
        }
    }
    cin>>ax>>ay>>bx>>by;
    wire();
    if(len!=0){
        len=g[bx][by]-2;
    }
    cout<<len;
    return 0;
}

Answer 1

有大佬可以帮帮我吗