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2014-10-01 12:19 阅读 3.9k

ACM一道题 poj3523 UVA1601双向广度优先BFS

我没有用双广,用的是紫书上说的把空格提出来重新建了一张图,调试了两天,实在找不出bug,第二组测试数据总是38而不是36。。。
哪位大神做过了这道题,跪求帮助啊!!
链接:http://poj.org/problem?id=3523

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1条回答 默认 最新

  • lxj1048580178 _forever_ 2015-01-17 06:57

    #include
    #include
    #include
    #include
    using namespace std;

    const int maxs = 20;
    const int maxn = 150; // 75% cells plus 2 fake nodes
    const int dx[]= {1,-1,0,0,0}; // 4 moves, plus "no move"
    const int dy[]= {0,0,1,-1,0};

    inline int ID(int a, int b, int c)
    {
    return (a<<16)|(b<<8)|c;
    }

    int s[3], t[3]; // starting/ending position of each ghost

    int deg[maxn], G[maxn][5]; // target cells for each move (including "no move")

    inline bool conflict(int a, int b, int a2, int b2)
    {
    return a2 == b2 || (a2 == b && b2 == a);
    }

    int d[maxn][maxn][maxn]; // distance from starting state

    int bfs()
    {
    queue q;
    memset(d, -1, sizeof(d));
    q.push(ID(s[0], s[1], s[2])); // starting node
    d[s[0]][s[1]][s[2]] = 0;
    while(!q.empty())
    {
    int u = q.front();
    q.pop();
    int a = (u>>16)&0xff, b = (u>>8)&0xff, c = u&0xff;
    if(a == t[0] && b == t[1] && c == t[2]) return d[a][b][c]; // solution found
    for(int i = 0; i < deg[a]; i++)
    {
    int a2 = G[a][i];
    for(int j = 0; j < deg[b]; j++)
    {
    int b2 = G[b][j];
    if(conflict(a, b, a2, b2)) continue;
    for(int k = 0; k < deg[c]; k++)
    {
    int c2 = G[c][k];
    if(conflict(a, c, a2, c2)) continue;
    if(conflict(b, c, b2, c2)) continue;
    if(d[a2][b2][c2] != -1) continue;
    d[a2][b2][c2] = d[a][b][c]+1;
    q.push(ID(a2, b2, c2));
    }
    }
    }
    }
    return -1;
    }

    int main()
    {
    int w, h, n;

    while(scanf("%d%d%d\n", &w, &h, &n) == 3 && n)
    {
        char maze[20][20];
        for(int i = 0; i < h; i++)
            fgets(maze[i], 20, stdin);
    
        // extract empty cells
        int cnt, x[maxn], y[maxn], id[maxs][maxs]; // cnt is the number of empty cells
        cnt = 0;
        for(int i = 0; i < h; i++)
            for(int j = 0; j < w; j++)
                if(maze[i][j] != '#')
                {
                    x[cnt] = i;
                    y[cnt] = j;
                    id[i][j] = cnt;
                    if(islower(maze[i][j])) s[maze[i][j] - 'a'] = cnt;
                    else if(isupper(maze[i][j])) t[maze[i][j] - 'A'] = cnt;
                    cnt++;
                }
    
        // build a graph of empty cells
        for(int i = 0; i < cnt; i++)
        {
            deg[i] = 0;
            for(int dir = 0; dir < 5; dir++)
            {
                int nx = x[i]+dx[dir], ny = y[i]+dy[dir];
                // "Outermost cells of a map are walls" means we don't need to check out-of-bound
                if(maze[nx][ny] != '#') G[i][deg[i]++] = id[nx][ny];
            }
        }
    
        // add fakes nodes so that in each case we have 3 ghosts. this makes the code shorter
        if(n <= 2)
        {
            deg[cnt] = 1;
            G[cnt][0] = cnt;
            s[2] = t[2] = cnt++;
        }
        if(n <= 1)
        {
            deg[cnt] = 1;
            G[cnt][0] = cnt;
            s[1] = t[1] = cnt++;
        }
    
        printf("%d\n", bfs());
    }
    return 0;
    

    }

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