hdu 1885 Key Task(bfs)

http://acm.hdu.edu.cn/showproblem.php?pid=1885

再贴一个链接http://blog.csdn.net/u013081425/article/details/21756237

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <stack>
#include <vector>
#include <queue>
#define LL long long
#define _LL __int64

using namespace std;
const int INF = 0x3f3f3f3f;

int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
struct node
{
    int x,y;
    int sta;
    int step;
};

int n,m;
int sx,sy;
char map[110][110];
int mark[110][110][(1<<4)+10];

queue <struct node>que;

int ans;

int small(char ch)
{
    if(ch == ‘b‘)
        return 0;
    else if(ch == ‘y‘)
        return 1;
    else if(ch == ‘r‘)
        return 2;
    else if(ch == ‘g‘)
        return 3;
}

int big(char ch)
{
    if(ch == ‘B‘)
        return 0;
    else if(ch == ‘Y‘)
        return 1;
    else if(ch == ‘R‘)
        return 2;
    else if(ch == ‘G‘)
        return 3;
}

void bfs()
{
    while(!que.empty()) que.pop();
    memset(mark,0,sizeof(mark));

    mark[sx][sy][0] = 1;
    que.push((struct node) {sx,sy,0,0});

    while(!que.empty())
    {
        struct node u = que.front();
        que.pop();
        if(map[u.x][u.y] == ‘X‘)
        {
            ans = u.step;
            return;
        }

        for(int d = 0; d < 4; d++)
        {
            int x = u.x + dir[d][0];
            int y = u.y + dir[d][1];
            int sta = u.sta;

            if(x < 1 || x > n || y < 1 || y > m) continue;

            if(map[x][y] == ‘.‘ || map[x][y] == ‘*‘ || map[x][y] == ‘X‘) //注意出口‘X’也要进队列
            {
                if(!mark[x][y][sta])
                {
                    mark[x][y][sta] = 1;
                    que.push((struct node){x,y,sta,u.step+1});
                }
            }
            else if(map[x][y] >= ‘a‘ && map[x][y] <= ‘z‘)
            {
                int add = small(map[x][y]);

                if((sta&(1<<add)) == 0)
                    sta += (1 << add);
                if(!mark[x][y][sta])
                {
                    mark[x][y][sta] = 1;
                    que.push((struct node){x,y,sta,u.step+1});
                }
            }
            else if(map[x][y] >= ‘A‘ && map[x][y] <= ‘Z‘)
            {
                int add = big(map[x][y]);
                if( (sta&(1<<add)) && !mark[x][y][sta] )
                {
                    mark[x][y][sta] = 1;
                    que.push((struct node){x,y,sta,u.step+1});
                }
            }
        }
    }
}


int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        if(n == 0 && m == 0) break;
        for(int i = 1; i <= n; i++)
        {
            scanf("%s",map[i]+1);
            for(int j = 1; j <= m; j++)
            {
                if(map[i][j] == ‘*‘)
                {
                    sx = i;
                    sy = j;
                }
            }
        }

        ans = -1;
        bfs();
        if(ans == -1)
            printf("The poor student is trapped!\n");
        else printf("Escape possible in %d steps.\n",ans);
    }
    return 0;
}

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hdu 1885 Key Task(bfs)

原文：http://blog.csdn.net/u013081425/article/details/24154577