UVA 1600 Patrol Robot （巡逻机器人）（bfs）

题意：从（1,1）走到（m，n），最多能连续穿越k个障碍，求最短路。

分析：obstacle队列记录当前点所穿越的障碍数，如果小于k可继续穿越障碍，否则不能，bfs即可。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 20 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int pic[MAXN][MAXN];
int vis[MAXN][MAXN];
int m, n, k;
bool judge(int x, int y){
    return x >= 1 && x <= m && y >= 1 && y <= n;
}
int bfs(){
    queue<int> x, y, step, obstacle;
    x.push(1);
    y.push(1);
    step.push(0);
    obstacle.push(0);
    vis[1][1] = 1;
    while(!x.empty()){
        int tmpx = x.front(); x.pop();
        int tmpy = y.front(); y.pop();
        int tmpstep = step.front(); step.pop();
        int tmpobstacle = obstacle.front(); obstacle.pop();
        for(int i = 0; i < 4; ++i){
            int tx = tmpx + dr[i];
            int ty = tmpy + dc[i];
            if(judge(tx, ty) && !vis[tx][ty]){
                if(tx == m && ty == n) return tmpstep + 1;
                if(pic[tx][ty] == 0){
                    vis[tx][ty] = 1;
                    x.push(tx);
                    y.push(ty);
                    step.push(tmpstep + 1);
                    obstacle.push(0);
                }
                else if(pic[tx][ty] == 1){
                    int nowobstacle = tmpobstacle + 1;
                    if(nowobstacle <= k){
                        x.push(tx);
                        y.push(ty);
                        step.push(tmpstep + 1);
                        obstacle.push(nowobstacle);
                    }
                }
            }
        }
    }
    return -1;
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        memset(pic, 0, sizeof pic);
        memset(vis, 0, sizeof vis);
        scanf("%d%d%d", &m, &n, &k);
        for(int i = 1; i <= m; ++i){
            for(int j = 1; j <= n; ++j){
                scanf("%d", &pic[i][j]);
            }
        }
        int ans = bfs();
        printf("%d\n", ans);
    }
    return 0;
}