洛谷P1825

題目鏈接

一個BFS，雖然洛谷難度標籤是提高，但是感覺還行。
很熟悉的傳送門，比較特殊的地方是需要剪枝（可能是因爲我題目刷少了，還沒遇到過BFS剪枝

AC代碼：

/*
 * @Author: hesorchen
 * @Date: 2020-04-14 10:33:26
 * @LastEditTime: 2020-06-24 23:36:50
 * @Link: https://hesorchen.github.io/
 */
#include <map>
#include <set>
#include <list>
#include <queue>
#include <deque>
#include <cmath>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl '\n'
#define PI acos(-1)
#define PB push_back
#define ll long long
#define INF 0x3f3f3f3f
#define mod 3123456777
#define pll pair<ll, ll>
#define lowbit(abcd) (abcd & (-abcd))
#define max(a, b) ((a > b) ? (a) : (b))
#define min(a, b) ((a < b) ? (a) : (b))

#define IOS                      \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0);
#define FRE                              \
    {                                    \
        freopen("in.txt", "r", stdin);   \
        freopen("out.txt", "w", stdout); \
    }

inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
//==============================================================================

bool vis[310][310];
bool vis2[310][310];
char mp[310][310];
ll mov[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
struct node
{
    ll x, y, step;
};
queue<node> q;

ll sx, sy, ex, ey;
vector<pll> vec[30];
ll ans = INF;
ll n, m;

void bfs()
{
    while (q.size())
    {
        node temp = q.front();
        q.pop();
        if (temp.x == ex && temp.y == ey)
        {
            ans = min(ans, temp.step);
            continue;
        }
        if (temp.step > ans) //剪枝 沒減tle四個點
            continue;
        for (int i = 0; i < 4; i++)
        {
            ll xx = temp.x + mov[i][0];
            ll yy = temp.y + mov[i][1];
            if (xx >= 1 && yy >= 1 && xx <= n && yy <= m && !vis[xx][yy] && mp[xx][yy] != '#' && vis2[xx][yy] <= 1)
            {
                if (mp[xx][yy] == '.' || mp[xx][yy] == '=')
                {
                    vis[xx][yy] = 1;
                    q.push(node{xx, yy, temp.step + 1});
                }
                else if (mp[xx][yy] <= 'Z' && mp[xx][yy] >= 'A')
                {
                    ll x1 = vec[mp[xx][yy] - 'A'][0].first;
                    ll y1 = vec[mp[xx][yy] - 'A'][0].second;
                    ll x2 = vec[mp[xx][yy] - 'A'][1].first;
                    ll y2 = vec[mp[xx][yy] - 'A'][1].second;

                    vis2[x1][y1]++;
                    vis2[x2][y2]++;
                    if (x1 == xx && y1 == yy)
                        q.push(node{x2, y2, temp.step + 1});
                    else if (x2 == xx && y2 == yy)
                        q.push(node{x1, y1, temp.step + 1});
                }
            }
        }
    }
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> mp[i] + 1;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
        {
            if (mp[i][j] == '=')
                ex = i, ey = j;
            if (mp[i][j] == '@')
                sx = i, sy = j;
            if (mp[i][j] <= 'Z' && mp[i][j] >= 'A')
                vec[mp[i][j] - 'A'].PB(make_pair(i, j));
        }
    vis[sx][sy] = 1;
    q.push(node{sx, sy, 0});
    bfs();
    cout << (ans == INF ? -1 : ans) << endl;
    return 0;
}