HDOJ 3085 Nightmare Ⅱ

题意:在不被ghost(Z)捉到的情况下,G和M是否能够相遇,人只能走有路的地方,而ghost可以穿墙,G的移动速度为1,Z移动速度为2,M速度为3

链接:http://acm.hdu.edu.cn/showproblem.php?pid=3085

思路:分别从G点和M点进行双向广度优先搜索,并且对于其中的每一个节点与Z点的Manhattan距离进行时间的计算。

注意点:无


以下为AC代码:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define RDI(a) scanf ( "%d", &a )
#define RDII(a, b) scanf ( "%d%d", &a, &b )
#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c );
#define RDS(s) scanf ( "%s", s );
#define REP(i,m,n) for ( int i = m; i <= n; i ++ )
#define DEP(i,m,n) for ( int i = m; i >= n; i -- )
#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )
#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
//const double pi = acos(-1);
const double eps = 1e-10;
const int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };
bool adj[2][1005][1005];
char str[1005][1005];
struct node{
    int x, y, cnt;
    node(){}
    node( int _x, int _y ) : x(_x), y(_y) {}
    node( int _x, int _y, int _cnt ) : x(_x), y(_y), cnt(_cnt) {}
};
int m, n;
node z[2];
queue q[2];
node y;
node g;
int cnt;
void init()
{
    clr ( adj, 0 );
    clr ( str, 'X' );
    cnt = 0;
    int num = 0;
    RDII ( m, n );
    REP ( i, 1, m ){
        RDS ( &str[i][1] );
        str[i][n+1] = 'X';
        REP ( j, 1, n ){
            if ( str[i][j] == 'Z' )
                z[num++] = node ( i, j );
            if ( str[i][j] == 'M' )
                y = node ( i, j );
            if ( str[i][j] == 'G' )
                g = node ( i, j );
        }
    }
}
void print()
{
    printf ( "\n" );
    REP ( k, 0, 1 ){
        REP ( i, 0, m + 1 ){
            REP ( j, 0, n + 1 ){
                printf ( "%d ", adj[k][i][j] );
            }
            printf ( "\n" );
        }
        printf ( "\n" );
    }
}
bool judge ( node tmp )
{
    if ( tmp.x < 0 || tmp.x > m || tmp.y < 0 || tmp.y > n || str[tmp.x][tmp.y] == 'X' )return false;
    REP ( i, 0, 1 ){
        //printf ( "%d %d %d %d\n", tmp.x, tmp.y, abs ( tmp.x - z[i].x ) + abs ( tmp.y - z[i].y ), 2 * cnt );
        if ( abs ( ( tmp.x - z[i].x ) ) + abs ( ( tmp.y - z[i].y ) ) <= 2 * cnt )
            return false;
    }
    return true;
}
bool bfs ( int k )
{
    int s = q[k].size();
    while ( s -- ){
        node tmp = q[k].front();
        //cout <<  k <<' ' << tmp.x << ' ' << tmp.y << endl;
        q[k].pop();
        if ( ! judge ( tmp ) )continue;
        REP ( i, 0, 3 ){
            int xi = tmp.x + dir[i][0];
            int yi = tmp.y + dir[i][1];
            if ( judge ( node ( xi, yi ) ) ){
                if ( adj[k][xi][yi] )continue;
                adj[k][xi][yi] = true;
                if ( adj[k^1][xi][yi] )return true;
                q[k].push ( node ( xi, yi ) );
            }
        }
    }
    return false;
}
bool solve()
{
    while ( ! q[0].empty() )q[0].pop();
    while ( ! q[1].empty() )q[1].pop();
    q[0].push ( y );
    q[1].push ( g );
    adj[0][y.x][y.y] = adj[1][g.x][g.y] = 1;
    while ( ( ! q[0].empty() ) || ( ! q[1].empty() ) ){
        cnt ++;
        //REP ( i, 1, 3 ){
            if ( bfs ( 0 ) )return true;
            if ( bfs ( 0 ) )return true;
            if ( bfs ( 0 ) )return true;
        //}
        if ( bfs ( 1 ) )return true;
        //cout << cnt << endl;
    }
    return false;
}
int main()
{
    int t;
    RDI(t);
    while ( t -- ){
        init();
        if ( solve() )
            printf ( "%d\n", cnt );
        else
            printf ( "-1\n" );
        //print();
    }
    return 0;
}


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