FZU Problem 2028 时空门问题

一开始是MLE,后来想到了用vector,化二维为一维,做了这一步优化后,这就是很基础的一个广搜了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
#include<vector>
using namespace std;
#define maxn 510
char maps[maxn][maxn];
vector<int>v[maxn*maxn];
int vis[maxn][maxn],go[4][2] = {1,0,0,1,0,-1,-1,0},n,m;
struct Pos
{
    int x,y,s;
}node;
bool ok(Pos a)
{
    return (a.x>=0 && a.x<n && a.y>=0 && a.y<m && maps[a.x][a.y] != ‘#‘ && !vis[a.x][a.y]);
}
int bfs(Pos a,Pos b)
{
    queue<Pos> que;
    memset(vis,0,sizeof(vis));
    while(!que.empty()) que.pop();
    vis[a.x][a.y] = 1;
    que.push(a);
    while(!que.empty())
    {
        Pos now = que.front();
        que.pop();
        if(now.x == b.x && now.y == b.y) return now.s;
        Pos nxt;
        for(int i = 0;i < 4;i++)
        {
            nxt.x = now.x + go[i][0];
            nxt.y = now.y + go[i][1];
            if(ok(nxt))
            {
                vis[nxt.x][nxt.y] = 1;
                nxt.s = now.s + 1;
                que.push(nxt);
            }
        }
        int s = now.x*m + now.y;
        for(int i = 0;i < v[s].size();i++)
        {
            nxt.x = v[s][i] / m;
            nxt.y = v[s][i] % m;
            if(ok(nxt))
            {
                vis[nxt.x][nxt.y] = 1;
                nxt.s = now.s + 1;
                que.push(nxt);
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        Pos node1,node2;
        for(int i = 0;i < n;i++)
        scanf("%s",maps[i]);
        int k,x,y,st,en;
        for(int i = 0;i < n;i++)
        {
            for(int j = 0;j < m;j++)
            {
                if(maps[i][j] == ‘s‘)
                    node1.x = i,node1.y = j,node1.s = 0;
                else if(maps[i][j] == ‘t‘)
                    node2.x = i,node2.y = j,node2.s = 0;
                st = i*m + j;
                v[st].clear();
                scanf("%d",&k);
                while(k--)
                {
                    scanf("%d%d",&x,&y);
                    x--,y--;
                    en = x*m + y;
                    v[st].push_back(en);
                }
            }
        }
        printf("%d\n",bfs(node1,node2));
    }
}