牛客 maze（最短路）

通过重新建图，跑一遍最短路可以获得答案。对于二维矩阵的形式，可以考虑转换成一维。

#include<bits/stdc++.h>
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<int,int> pll;
const int N=3e5+10;
int h[N],ne[N],e[N],w[N],idx;
char s[500][500];
int st,ed;
int id[5000][5000];
int t;
int dis[N];
bool vis[N];
int n,m,qi;
void add(int a,int b,int c){
    e[idx]=b,ne[idx]=h[a],w[idx]=c,h[a]=idx++;
}
void dij(){
    priority_queue<pll,vector<pll>,greater<pll> > q ;
    memset(dis,0x3f,sizeof dis);
    memset(vis,0,sizeof vis);
    dis[st]=0;
    q.push({0,st});
    while(q.size()){
        auto t=q.top();
        q.pop();
        int pos=t.se;
        if(pos==ed)
            break;
        if(vis[pos])
            continue;
        vis[pos]=1;
        for(int i=h[pos];i!=-1;i=ne[i]){
            int j=e[i];
            if(dis[j]>dis[pos]+w[i]){
                dis[j]=dis[pos]+w[i];
                q.push({dis[j],j});
            }
        }
    }
}
void build(){
    int i,j;
    memset(h,-1,sizeof h);
    int dx[]={-1,0,1,0},dy[]={0,1,0,-1};
    for(i=1;i<=n;i++){
        for(j=1;j<=m;j++){
            for(int u=0;u<4;u++){
                int x=i+dx[u],y=j+dy[u];
                if(x<=0||x>n||y<=0||y>m)
                    continue;
                if(s[i][j]=='#'||s[x][y]=='#')
                    continue;
                int a=id[i][j],b=id[x][y];
                add(a,b,1);
            }
        }
    }
}
int main(){
    int t;
    while(cin>>n>>m>>qi){
        int i,j;
        idx=0;
        int cnt=0;
        for(i=1;i<=n;i++){
            scanf("%s",s[i]+1);
            for(j=1;j<=m;j++){
                id[i][j]=++cnt;
                if(s[i][j]=='S')
                    st=id[i][j];
                if(s[i][j]=='T')
                    ed=id[i][j];
            }
        }
        build();
        while(qi--){
            int x1,y1,x2,y2;
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            if(s[x1+1][y1+1]=='#'||s[x2+1][y2+1]=='#')
                continue;
            int a=id[x1+1][y1+1],b=id[x2+1][y2+1];
            add(a,b,3);
        }
        dij();
        if(dis[ed]==0x3f3f3f3f)
            cout<<-1<<endl;
        else
            cout<<dis[ed]<<endl;
    }
}