UVA - 1600 Patrol Robot（BFS 三維判重）

題目大意：m * n 的地圖，機器人要從（0, 0）走到（m, n），上下左右移動，0 可走 1 爲障礙，k 表示可以連續通過的障礙數，輸出最小步數，到達不了輸出 -1
解題思路：vis 多一維判斷已經通過的障礙，別的就是套 BFS

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<string.h>
#include<string>
#include<queue>
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
const int INF = 0x3f3f3f3f;
const int NINF = -INF -1;
const int MAXN = 200+10;
using namespace std;
int T, k, n, m;
int map[MAXN][MAXN];
bool vis[MAXN][MAXN][MAXN];
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, -1, 0, 1};
struct point {
    int x, y;
    int cnt;
    int k;
};
int bfs() {
    queue<point> q;
    point now, p;
    now.x = now. y = now.cnt = now.k = 0;
    q.push(now);
    vis[0][0][0] = 1;
    while (!q.empty()) {
        now = q.front();
        q.pop();
//printf("%d %d %d\n", now.x, now.y, now.k);
        if (now.x == n-1 && now.y == m-1) return now.cnt;
        for (int i = 0; i < 4; i++) {
            p = now;
            p.x += dx[i];
            p.y += dy[i];
            p.cnt++;
            if (map[p.x][p.y]) p.k++;
            else p.k = 0;
            if (p.x < 0 || p.x >= n || p.y < 0 || p.y >= m) continue;
            if (!vis[p.x][p.y][p.k] && p.k <= k) {
//printf("==%d %d %d\n", p.x, p.y, p.k);
                q.push(p);
                vis[p.x][p.y][p.k] = 1;
            }
        }
    }
    return -1;
}
int main() {
    scanf("%d", &T);
    while (T--) {
        memset(vis, 0, sizeof(vis));
        scanf("%d%d%d", &n, &m, &k);
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                scanf("%d", &map[i][j]);
        printf("%d\n", bfs());

    }

    return 0;
}