POJ 1475 Pushing Boxes

标签：style   class   code   com   string   os   

搜索这种东西只要写之前规划得当还是蛮顺手的。。

mark[x1][y1][x2][y2]表示小人在(x1,y1) 盒子在(x2,y2)这种状态是否到过。

剩下的就是优先队列 + bfs 了，另外开一个栈记录前驱以输出路径。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <cmath>
#include <stack>
#include <map>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-8)
#define LL long long
#define ULL unsigned long long
#define _LL __int64
#define _INF 0x3f3f3f3f
#define Mod 9999991

using namespace std;

char Map[24][24];

int mark[22][22][22][22];

struct N
{
    int x1,x2,y1,y2;
};

struct Q
{
    int ans,x1,y1,x2,y2,pre;
    int ans2,site;
    bool operator < (const Q &a)const{
        return (a.ans2 < ans2 || (a.ans2 == ans2 && a.ans < ans));
    }
}q[200000];

bool Judge(Q f)
{
    if(abs(f.x1-f.x2) + abs(f.y1-f.y2) <= 1)
        return true;
    return false;
}

int jx[] = {-1, 1, 0, 0};
int jy[] = { 0, 0,-1, 1};

void Out(int site,Q f)
{
    if(site == -1)
        return ;

    Out(q[site].pre,q[site]);

    Q t = q[site];
    
    if(t.x1 == f.x1)
    {
        if(t.y1+1 == f.y1)
        {
            if(t.x2 == f.x2 && t.y2+1 == f.y2)
            {
                printf("E");
            }
            else
            {
                printf("e");
            }
        }
        else if(t.y1-1 == f.y1)
        {
            if(t.x2 == f.x2 && t.y2-1 == f.y2)
            {
                printf("W");
            }
            else
            {
                printf("w");
            }
        }
    }
    else if(t.y1 == f.y1)
    {
        if(t.x1+1 == f.x1)
        {
            if(t.y2 == f.y2 && t.x2+1 == f.x2)
            {
                printf("S");
            }
            else
            {
                printf("s");
            }
        }
        else if(t.x1-1 == f.x1)
        {
            if(t.y2 == f.y2 && t.x2-1 == f.x2)
            {
                printf("N");
            }
            else
            {
                printf("n");
            }
        }
    }
}

void bfs(N s,int n,int m)
{
    memset(mark,-1,sizeof(mark));

    mark[s.x1][s.y1][s.x2][s.y2] = 0;
    priority_queue<Q> pq;

    Q f,t;

    int S = 0,E = 0;

    f.ans = 0;
    f.ans2 = 0;
    f.pre = -1;
    f.x1 = s.x1;
    f.x2 = s.x2;
    f.y1 = s.y1;
    f.y2 = s.y2;
    f.site = 0;
    q[E++] = f;
    pq.push(f);

    while(pq.empty() == false)
    {
        f = pq.top();
        pq.pop();

        if(Map[f.x2][f.y2] == 'T')
        {
            Out(f.pre,f);
            puts("");
            return ;
        }

        t.pre = f.site;
        t.ans = f.ans+1;
        for(int i = 0 ;i < 4; ++i)
        {
            t.x1 = f.x1 + jx[i];
            t.y1 = f.y1 + jy[i];
            t.x2 = f.x2;
            t.y2 = f.y2;

            if(1 <= t.x1 && t.x1 <= n && 1 <= t.y1 && t.y1 <= m && Map[t.x1][t.y1] != '#' && -1 == mark[t.x1][t.y1][t.x2][t.y2])
            {
                if(t.x1 != t.x2 || t.y1 != t.y2)
                {
                    mark[t.x1][t.y1][t.x2][t.y2] = t.ans;
                    t.site = E;
                    t.ans2 = f.ans2;
                    q[E++] = t;
                    pq.push(t);
                }
                else
                {
                    t.x2 = f.x2 + jx[i];
                    t.y2 = f.y2 + jy[i];

                     if(1 <= t.x2 && t.x2 <= n && 1 <= t.y2 && t.y2 <= m && Map[t.x2][t.y2] != '#' && -1 == mark[t.x1][t.y1][t.x2][t.y2])
                     {
                         t.ans2 = f.ans2+1;
                         t.site = E;
                         q[E++] = t;
                         pq.push(t);
                         mark[t.x1][t.y1][t.x2][t.y2] = t.ans;
                     }
                }
            }
        }
    }
    printf("Impossible.\n");
}

int main()
{
    int i,j,n,m;

    int icase = 1;

    while(scanf("%d %d",&n,&m) && (n||m))
    {
        for(i = 1;i <= n;++i)
            scanf("%s",Map[i]+1);

        N s;

        for(i = 1;i <= n;++i)
        {
            for(j = 1;j <= m; ++j)
            {
                if(Map[i][j] == 'S')
                    s.x1 = i,s.y1 = j;
                else if(Map[i][j] == 'B')
                    s.x2 = i,s.y2 = j;
            }
        }
        printf("Maze #%d\n",icase++);
        bfs(s,n,m);
        puts("");
    }
    return 0;
}

POJ 1475 Pushing Boxes,布布扣,bubuko.com

POJ 1475 Pushing Boxes

标签：style   class   code   com   string   os   

原文地址：http://blog.csdn.net/zmx354/article/details/33346705