AOJ 0558 Cheese 《挑战程序设计竞赛(第2版)》练习题答案 � 码农场


2.1 最基础的"穷竭搜索" 广度优先搜索
在H * W的地图上有N个奶酪工厂,分别生产硬度为1-N的奶酪。有一只吃货老鼠准备从老鼠洞出发吃遍每一个工厂的奶酪。老鼠有一个体力值,初始时为1,每吃一个工厂的奶酪体力值增加1(每个工厂只能吃一次),且老鼠只能吃硬度不大于当前体力值的奶酪。 
老鼠从当前格走到相邻的无障碍物的格(上下左右)需要时间1单位,有障碍物的格不能走。走到工厂上时即可吃到该工厂的奶酪,吃奶酪时间不计。问吃遍所有奶酪最少用时。 
输入:第一行三个整数H(1 <= H <= 1000)、W(1 <= W <=1000)、N(1 <= N <= 9),之后H行W列为地图, "."为空地, "X"为障碍物,"S"为老鼠洞, 1-N代表硬度为1-N的奶酪的工厂。(中文翻译参考了http://bbs.byr.cn/#!article/ACM_ICPC/73337?au=Milrivel)
10 10 9
.X...X.S.X
6..5X..X1X
...XXXX..X
X..9X...X.
8.X2X..X3X
...XX.X4..
XX....7X..
X..X..XX..
X...X.XX..
..X.......
思路:吃货必须按照工厂N值从小到大的顺序吃,否则体力不济。所以这个题目其实就是求按顺序遍历地图上12345……这几个点的最短路径。说到最短路径,当然就是bfs了。

分析:简单迷宫问题。不同的是,老鼠需要按1-N 的顺序把奶酪吃完。用广度优先搜索很容易求出起点到终点的最小步数。初始时,求起点到硬度值为 1 的奶酪的最小步数;接着将起点重置为此位置,继续求此位置到达硬度值为 2 的奶酪;如此类推。因此这里只需做N 次广度优先搜索,并累计其值即可。

int w, h, n;
char map[1024][1024];
// 各点到当前工厂的距离
int d[1024][1024];
const int direction[4][2] = {
    { -1, 0 },
    { 1, 0 },
    { 0, -1 },
    { 0, 1 },
};
int factory[16][2];
typedef pair<intint> P;
// Parameter: const int & sx 起点x
// Parameter: const int & sy 起点y
// Parameter: const int & gx 终点x
// Parameter: const int & gy 终点y
int bfs(const int& sx, const int& sy, const int& gx, const int& gy)
{
    //memset(d, -1, sizeof(d));
    for (int i = 0; i < h; ++i)
    {
        for (int j = 0; j < w; ++j)
        {
            d[j][i] = -1;
        }
    }
    queue<P> que;
    que.push(P(sx, sy));
    d[sx][sy] = 0;
    while (que.size())
    {
        const P p = que.front(); que.pop();
        // 如果是终点就结束
        if (p.first == gx && p.second == gy)
        {
            break;
        }
        // 四方向漫游
        for (int i = 0; i < 4; ++i)
        {
            int nx = p.first + direction[i][0];
            int ny = p.second + direction[i][1];
            // 是否可以移动,并且该点没有访问过
            if (0 <= nx && nx < w && 0 <= ny && ny < h && map[nx][ny] != 'X' && d[nx][ny] == -1)
            {
                que.push(P(nx, ny));
                d[nx][ny] = d[p.first][p.second] + 1;
            }
        }
    }
    return d[gx][gy];
}
int main(int argc, char *argv[])
{
    cin >> h >> w >> n;
    for (int i = 0; i < h; ++i)
    {
        for (int j = 0; j < w; ++j)
        {
            cin >> map[j][i];
        }
    }
    for (int i = 0; i < h; ++i)
    {
        for (int j = 0; j < w; ++j)
        {
            if (map[j][i] == 'S')
            {
                factory[0][0] = j;
                factory[0][1] = i;
                map[j][i] = '.';
            }
            else if (isdigit(map[j][i]))
            {
                int index = map[j][i] - '0';
                factory[index][0] = j;
                factory[index][1] = i;
            }
        }
    }
    int step = 0;
    for (int i = 0; i < n; ++i)
    {
        // 按顺序吃遍中华
        step += bfs(factory[i][0], factory[i][1], factory[i + 1][0], factory[i + 1][1]);
    }
    cout << step << endl;
    return 0;
}
Also check http://www.cnblogs.com/7hat/p/3594076.html
int H, W, N;
16 char maze[MAX_H][MAX_W + 1];
17 
18 int sx, sy;     //start
19 int d[MAX_H][MAX_W];     //steps
20 
21 const int dx[4] = {-1, 1, 0, 0};
22 const int dy[4] = {0, 0, -1, 1};
23 
24 int bfs(char c){
25     //init
26     for(int i = 0; i < H; i ++){
27         fill(d[i], d[i] + W, INF);
28     }
29     d[sx][sy] = 0;
30     queue<P> que;
31     que.push(P(sx, sy));
32 
33     while(!que.empty()){
34         P p = que.front();
35         que.pop();
36         //arrive
37         if(maze[p.first][p.second] == c){
38         //reset
39         sx = p.first;
40         sy = p.second;
41         break;
42         }
43 
44         for(int i = 0; i < 4; i ++){
45             int nx = p.first + dx[i], ny = p.second + dy[i];
46 
47             if(0 <= nx && nx < H && 0 <= ny && ny < W && maze[nx][ny] != 'X' && d[nx][ny] == INF){
48                 que.push(P(nx, ny));
49                 d[nx][ny] = d[p.first][p.second] + 1;
50             }
51         }
52     }
53     return d[sx][sy];
54 }
55 
56 void solve(){
57     //start
58     for(int i = 0; i < H; i ++){
59         for(int j = 0; j < W; j ++){
60             if(maze[i][j] == 'S'){
61                 sx = i;
62                 sy = j;
63                 break;
64             }
65         }
66     }
67     //bfs for 1-N
68     int ans = 0;
69     for(int i = 1; i <= N; i ++){
70         ans += bfs('0' + i);
71     }
72     printf("%d\n", ans);
73 }
74 
75 int main(int argc, char const *argv[]){
76 
77     scanf("%d %d %d", &H, &W, &N);
78     for(int i = 0; i < H; i ++){
79         scanf("%s", maze[i]);
80     }
81     solve();
82 
83     return 0;
84 }
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