穿越栅栏 Overfencing

题解区里都是一次性走两步

这里我就说一点

按普通(每次走一步)的来最后\(+1\)再除以\(2\)就行了

另外输入时加个时间限制 （\(q++ if(q>=w)break;\)）

/*
ID:death_r2
TASK:maze1
LANG:C++
*/
#include <queue>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define reg register int
#define isdigit(x) ('0' <= x&&x <= '9')
template<typename T>
inline T Read(T Type)
{
    T x = 0,f = 1;
    char a = getchar();
    while(!isdigit(a)) {if(a == '-') f = -1;a = getchar();}
    while(isdigit(a)) {x = (x << 1) + (x << 3) + (a ^ '0');a = getchar();}
    return x * f;
}
string mp[210];
bool vis[3][210][210];
int cnt,ans,d[3][210][210],dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}},w,h;
pair<int,int> s[3];
#define pii_pii pair<int,pair<int,int> >
inline void bfs(pair<int,int> s,int index)
{
    priority_queue<pii_pii,vector<pii_pii >,greater<pii_pii > > q;
    q.push(make_pair(0,s));
    vis[index][s.first][s.second] = 1;
    while(!q.empty())
    {
        pii_pii it = q.top();q.pop();
        for(reg i = 0;i < 4;i++)
        {
            int x = it.second.first + dir[i][0];
            int y = it.second.second + dir[i][1];
            if(x < 1||x > h||y < 0||y >= w||mp[x][y] != ' '||vis[index][x][y]) continue;
            d[index][x][y] = d[index][it.second.first][it.second.second] + 1;
            vis[index][x][y] = 1;
            q.push(make_pair(d[index][x][y],make_pair(x,y)));
        }
    }
}
int main()
{
//    freopen("maze1.in","r",stdin);
//    freopen("maze1.out","w",stdout);
    w = Read(1) << 1 ^ 1,h = Read(1) << 1 ^ 1;
    for(reg i = 1;i <= h;i++)
    {
        char x = getchar();
        while(x != ' '&&x != '|'&&x != '+'&&x != '-') x = getchar();
        int q = 0;
        while(x != '\n')
        {
            ++q;
            mp[i] += x;
            x = getchar();
            if(q >= w) break;
        }
    }
    for(reg i = 1;i <= h;i++)
        while(mp[i].size() < w) mp[i] += ' ';
    for(reg i = 0;i < w;i++)
    {
        if(cnt == 2) break;
        if(mp[1][i] == ' ') s[++cnt] = make_pair(1,i);
        if(mp[h][i] == ' ') s[++cnt] = make_pair(h,i);
    }
    for(reg i = 1;i <= h;i++)
    {
        if(cnt == 2) break;
        if(mp[i][0] == ' ') s[++cnt] = make_pair(i,0);
        if(mp[i][w - 1] == ' ') s[++cnt] = make_pair(i,w - 1);
    }
    bfs(s[1],1),bfs(s[2],2);
    for(reg i = 1;i <= h;i++)
        for(reg j = 0;j < w;j++)
            ans = max(ans,min((d[1][i][j] + 1) / 2,(d[2][i][j] + 1) / 2));
    printf("%d\n",ans);
    return 0;
}

来源：https://www.cnblogs.com/resftlmuttmotw/p/11965579.html