bfs UESTC 381 Knight and Rook

http://acm.uestc.edu.cn/#/problem/show/381

题目大意：给你两个棋子：车、马，再给你一个n*m的网格，从s出发到t，你可以选择车或者选择马开始走，图中有一些障碍物，该障碍物是不能走的，走的图中有换一次棋子的机会，问最少需要几次能从s走到t？

思路：bfs来4次就好了。两次记录从s->t，两次是t->s。然后暴力一下就出来了，复杂度为4*n*m*log

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n")
const int maxn = 100 + 5;
const int inf = 0x3f3f3f3f;
char ch[maxn][maxn];
int n, m;
pair<int, int> s, t;
int rook[maxn][maxn], knight[maxn][maxn];
int r1[maxn][maxn], k1[maxn][maxn];
int r2[maxn][maxn], k2[maxn][maxn];
int dx[] = {-2, -2, -1, 1, 2, 2, 1, -1};
int dy[] = {-1, 1, 2, 2, 1, -1, -2, -2};

int bfs1(int a, int b){
    memset(rook, 0x3f, sizeof(rook));
    queue<pair<int, int> > que;
    que.push(mk(a, b));
    rook[a][b] = 0;
    while (!que.empty()){
        pair<int, int> p = que.front(); que.pop();
        int x = p.fi, y = p.se;
        ///if (flag && x == s.fi && y == s.se) break;
        int lb = 0, rb = 0;
        for (int i = x - 1; i > 0; i--)
            if (ch[i][y] == '#') {lb = i; break;}
        for (int i = x; i <= n; i++)
            if (ch[i][y] == '#') {rb = i; break;}
        if (rb == 0) rb = n + 1;
        ///lb和rb都是不能走的
        for (int i = lb + 1; i < rb; i++){
            if (i != x && rook[i][y] > rook[x][y] + 1){
                rook[i][y] = rook[x][y] + 1;
                que.push(mk(i, y));
            }
        }
        lb = 0, rb = 0;
        for (int i = y - 1; i > 0; i--)
            if (ch[x][i] == '#') {lb = i; break;}
        for (int i = y; i <= m; i++)
            if (ch[x][i] == '#') {rb = i; break;}
        if (rb == 0) rb = m + 1;
        for (int i = lb + 1; i < rb; i++){
            if (i != y && rook[x][i] > rook[x][y] + 1){
                rook[x][i] = rook[x][y] + 1;
                que.push(mk(x, i));
            }
        }
    }
    return rook[s.fi][s.se];
}

int bfs2(int a, int b){
    memset(knight, 0x3f, sizeof(knight));
    queue<pair<int, int> > que;
    que.push(mk(a, b));
    knight[a][b] = 0;
    while (!que.empty()){
        pair<int, int> p = que.front(); que.pop();
        int x = p.fi, y = p.se;
        ///if (flag && x == s.fi && y == s.se) break;
        for (int i = 0; i < 8; i++){
            int nx = x + dx[i], ny = y + dy[i];
            if (nx <= 0 || ny <= 0 || nx > n || ny > m) continue;
            if (ch[nx][ny] == '#') continue;
            if (knight[nx][ny] > knight[x][y] + 1){
                knight[nx][ny] = knight[x][y] + 1;
                que.push(mk(nx, ny));
            }
        }
    }
    return rook[s.fi][s.se];
}

int solve(){
    //printf("k = %d r = %d\n", k[s.fi][s.se], r[s.fi][s.se]);
    //int mini = min(r2[t.fi][t.se], k2[t.fi][t.se]);
    int mini = inf;
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= m; j++){
            if (ch[i][j] == '#') continue;
            if (r1[i][j] < inf && k2[i][j] < inf) mini = min(mini, r1[i][j] + k2[i][j]);
            if (k1[i][j] < inf && r2[i][j] < inf) mini = min(mini, k1[i][j] + r2[i][j]);
        }
    }
    if (mini == inf) return -1;
    return mini;
}

int main(){
    int T; cin >> T;
    for (int kase = 1; kase <= T; kase++){
        cin >> n >> m;
        for (int i = 1; i <= n; i++){
            scanf("%s", ch[i] + 1);
            for (int j = 1; j <= m; j++){
                if (ch[i][j] == 's') s = mk(i, j);
                if (ch[i][j] == 't') t = mk(i, j);
            }
        }
        memset(r1, inf, sizeof(r1)); memset(k1, inf, sizeof(k1));
        memset(r2, inf, sizeof(r2)); memset(k2, inf, sizeof(k2));
        bfs1(t.fi, t.se); bfs2(t.fi, t.se);
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                r1[i][j] = rook[i][j], k1[i][j] = knight[i][j];
        bfs1(s.fi, s.se); bfs2(s.fi, s.se);
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                r2[i][j] = rook[i][j], k2[i][j] = knight[i][j];
        printf("Case #%d: %d\n", kase, solve());
    }
    return 0;
}