poj1101 The Game

topic:

A game to a board, there are blank and can not walk the grid, each given a board sets the start and end, determine which two can not be together. The space and peripherals can go. If Internet connectivity, find the minimum number of bends.

---

This problem is something that many pit

• The length of each board, the width is reversed
• Start, end of the horizontal and vertical coordinates are reversed

• Each board to wrap the output

  Thinking

• The BFS (Priority Queue)
• Enter the very pit, Refer to code
• Recording the number of turning sum array for each state

• Direction enumerate four directions, you can go outside, so pay attention to the border, because it is seeking the number of turn, so to record the last judgment

  Code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
int n,m,sx,sy,ex,ey;
bool map[80][80];
int sum[4][160][160],T1,T2,ans;
int dir[4][2]={{-1,0},{0,-1},{1,0},{0,1}};
 struct ss
{
    int x,y,d,num;
    friend bool operator < (ss a,ss b)
    {
        return a.num>b.num;
    }
};
priority_queue<ss>q;

 void bfs()
{
    memset(sum,0xf,sizeof(sum));
    while (!q.empty()) q.pop();
    ss res,tmp;
    res.x=sx; res.y=sy; res.num=1; res.d=0; q.push(res);
    res.d=1; q.push(res);
    res.d=2; q.push(res);
    res.d=3; q.push(res);
    sum[0][sx][sy]=sum[1][sx][sy]=sum[2][sx][sy]=sum[3][sx][sy]=1;
    while (!q.empty())
    {
        res=q.top(); q.pop();
        if (res.x==ex&&res.y==ey)
        {
            ans=res.num;
            return;
        }
        if (res.num>sum[res.d][res.x][res.y]) continue;
        int num1,x,y;
        for (int i=0; i<4; i++)
        {
            if (i==res.d) num1=res.num;
            else num1=res.num+1;
            x=res.x+dir[i][0]; y=res.y+dir[i][1];
            while (x>=0&&x<=n+1&&y>=0&&y<=m+1&&!map[x][y])
            {
                if (num1<sum[i][x][y])
                {
                    sum[i][x][y]=num1;
                    tmp.x=x; tmp.y=y; tmp.num=num1; tmp.d=i;
                    q.push(tmp);
                }
                x=x+dir[i][0]; y=y+dir[i][1];
                if (map[x][y]) break;
            }
            if (x==ex&&y==ey)
                if (num1<sum[i][x][y])
                {
                    sum[i][x][y]=num1;
                    tmp.x=x; tmp.y=y; tmp.num=num1; tmp.d=i;
                    q.push(tmp);
                }
        }
    }
}
 void init()
{
    memset(map,0,sizeof(map));
    char c;
    int j,T2=0;
    while(scanf("%c",&c)==1)
    {
        if(c=='\n')break;
    }
    for(int i=1;i<=n;i++)
    {
        j=0;
        while(scanf("%c",&c)==1)
        {
            j++;
            if(c=='X') map[i][j]=1;
            if(c=='\n') break;
        }
    }
    printf("Board #%d:\n",++T1);
    while(scanf("%d%d%d%d",&sy,&sx,&ey,&ex)==4&&sx)//横纵坐标是反的
    {
        printf("Pair %d:",++T2);
        ans=0;
        bfs();
        if(ans) printf(" %d segments.\n",ans);
        else printf(" impossible.\n");
    }
}

 int main()
{
    while (scanf("%d%d",&m,&n))
    {
        if (!m&&!n) break;
        init(); printf("\n");//坑点,千万注意
    }
    return 0;
}