poj 2049 Finding Nemo 建迷宫预处理+优先队列bfs

Finding Nemo

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 6767		Accepted: 1546

Description

Nemo is a naughty boy. One day he went into the deep sea all by himself. Unfortunately, he became lost and couldn't find his way home. Therefore, he sent a signal to his father, Marlin, to ask for help. 
After checking the map, Marlin found that the sea is like a labyrinth with walls and doors. All the walls are parallel to the X-axis or to the Y-axis. The thickness of the walls are assumed to be zero. 
All the doors are opened on the walls and have a length of 1. Marlin cannot go through a wall unless there is a door on the wall. Because going through a door is dangerous (there may be some virulent medusas near the doors), Marlin wants to go through as few doors as he could to find Nemo. 
Figure-1 shows an example of the labyrinth and the path Marlin went through to find Nemo. 

We assume Marlin's initial position is at (0, 0). Given the position of Nemo and the configuration of walls and doors, please write a program to calculate the minimum number of doors Marlin has to go through in order to reach Nemo.

Input

The input consists of several test cases. Each test case is started by two non-negative integers M and N. M represents the number of walls in the labyrinth and N represents the number of doors. 
Then follow M lines, each containing four integers that describe a wall in the following format: 
x y d t 
(x, y) indicates the lower-left point of the wall, d is the direction of the wall -- 0 means it's parallel to the X-axis and 1 means that it's parallel to the Y-axis, and t gives the length of the wall. 
The coordinates of two ends of any wall will be in the range of [1,199]. 
Then there are N lines that give the description of the doors: 
x y d 
x, y, d have the same meaning as the walls. As the doors have fixed length of 1, t is omitted. 
The last line of each case contains two positive float numbers: 
f1 f2 
(f1, f2) gives the position of Nemo. And it will not lie within any wall or door. 
A test case of M = -1 and N = -1 indicates the end of input, and should not be processed.

Output

For each test case, in a separate line, please output the minimum number of doors Marlin has to go through in order to rescue his son. If he can't reach Nemo, output -1.

Sample Input

8 9
1 1 1 3
2 1 1 3
3 1 1 3
4 1 1 3
1 1 0 3
1 2 0 3
1 3 0 3
1 4 0 3
2 1 1
2 2 1
2 3 1
3 1 1
3 2 1
3 3 1
1 2 0
3 3 0
4 3 1
1.5 1.5
4 0
1 1 0 1
1 1 1 1
2 1 1 1
1 2 0 1
1.5 1.7
-1 -1

Sample Output

5
-1

题目大意：有一个迷宫，在迷宫中有墙与门
有m道墙，每一道墙表示为(x,y,d,t）
x,y表示墙的起始坐标
d为0即向右t个单位，都是墙
d为1即向上t个单位，都是墙
有n道门，每一道门表示为(x,y,d)
x,y表示门的起始坐标
d为0即向右一个单位表示门
d为1即向上一个单位表示门
再给出你起点的位置(f1,f2)，并保证这个点的位置不会再墙或者门中，为起点到(0,0)最少要穿过多少条门

解题思路：将坐标系看成网络 则墙和门可以看做网络上的边 每个区域对应四条边

感想：这题有点坑 第一是Nemo 可能不在迷宫内 那样直接输出0 第二是用C++提交为WA 用G++提交AC 应该是和编译器的输入有关吧  建议大家在输入量不是很很大的情况下  都用G++ 因为G++比较稳定

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#define maxn 210
using namespace std;

int n,m,maxx,maxy,ans,flag,tstep;
int ex,ey;
int mp[maxn][maxn];
int mpx[maxn][maxn];
int mpy[maxn][maxn];
int vis[maxn][maxn];
int sta[maxn*maxn];
int dx[4]={0,0,-1,1};    // up down left right
int dy[4]={1,-1,0,0};
struct Node
{
    int x,y;
    int step;
    friend bool operator <(Node x1,Node x2)    // 优先队列 按经过的门的次数排序
    {
        return x1.step>x2.step;
    }
}cur,now;
priority_queue<Node>q;

bool isok(int u,int v,int dd)           // 判断是否能走  要注意方向不同时 要判断不同额边
{
    int i,j;
    tstep=0;
    if(u+dx[dd]>maxx+1||u+dx[dd]<0||v+dy[dd]>maxy+1||v+dy[dd]<0||vis[u+dx[dd]][v+dy[dd]]) return false;
    if(dd==0)
    {
        if(mpx[u][v+dy[dd]]==2) return false;
        else if(mpx[u][v+dy[dd]]) tstep=1;
    }
    else if(dd==1)
    {
        if(mpx[u][v]==2) return false;
        else if(mpx[u][v]) tstep=1;
    }
    else if(dd==2)
    {
        if(mpy[u][v]==2) return false;
        else if(mpy[u][v]) tstep=1;
    }
    else
    {
        if(mpy[u+dx[dd]][v]==2) return false;
        else if(mpy[u+dx[dd]][v]) tstep=1;
    }
    return true;
}
bool bfs()
{
    int i,xx,yy,cnt;
    while(!q.empty()) q.pop();
    memset(vis,0,sizeof(vis));
    cur.x=0;
    cur.y=0;
    cur.step=0;
    vis[0][0]=1;
    q.push(cur);
    while(!q.empty())
    {
        now=q.top();
        xx=now.x;
        yy=now.y;
        cnt=now.step;
        if(xx==ex&&yy==ey)
        {
            ans=cnt;
            return true;
        }
        for(i=0;i<4;i++)
        {
            if(isok(xx,yy,i))
            {
                cur.x=xx+dx[i];
                cur.y=yy+dy[i];
                cur.step=cnt+tstep;        // 经过门则加1 不经过则加0
                q.push(cur);
                vis[cur.x][cur.y]=1;
            }
        }
        q.pop();
    }
    return false;
}
int main()
{
    int i,j;
    int tx,ty,d,l;
    double t1,t2;
    char c;
    while(scanf("%d%d",&n,&m))
    {
        if(n==-1&&m==-1) break;
        memset(mpx,0,sizeof(mpx));
        memset(mpy,0,sizeof(mpy));
        maxx=maxy=-1;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d%d",&tx,&ty,&d,&l);
            if(d)
            {
                for(j=ty;j<ty+l;j++)
                {
                    mpy[tx][j]=2;                // 墙
                }
                if(maxy<ty+l-1) maxy=ty+l-1;
            }
            else
            {
                for(j=tx;j<tx+l;j++)
                {
                    mpx[j][ty]=2;
                }
                if(maxx<tx+l-1) maxx=tx+l-1;
            }
        }
        for(i=1;i<=m;i++)
        {
            scanf("%d%d%d",&tx,&ty,&d);
            if(d)
            {
                for(j=ty;j<ty+1;j++)
                {
                    mpy[tx][j]=1;                // 门
                }
            }
            else
            {
                for(j=tx;j<tx+1;j++)
                {
                    mpx[j][ty]=1;
                }
            }
        }
        scanf("%lf%lf",&t1,&t2);
        ex=int(t1);
        ey=int(t2);
        flag=ans=0;
        if(ex>maxx+1||ey>maxy+1||ex<0||ey<0) flag=1;
        if(flag||bfs()) printf("%d\n",ans);
        else  printf("-1\n");
    }
    return 0;
}

附：其实这题poj上的数据有些是不严谨的比如  

4 1
3 2 1 1
1 3 0 2
1 1 0 2
1 1 1 2
2 1 1
1.5 1.5

这个数据答案应该是0  而不是1   这个是小峰峰想到的  在此赞一个  O(∩_∩)O

这里贴一份严谨代码吧  不过在poj上过不了 因为他的数据答案是错的

就是把标记数组换成三维  搜索的时候多搜一层就OK了

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#define maxn 210
using namespace std;

int n,m,maxx,maxy,ans,flag,tstep;
int ex,ey;
int mp[maxn][maxn];
int mpx[maxn][maxn];
int mpy[maxn][maxn];
int vis[maxn][maxn][4];
int sta[maxn*maxn];
int dx[4]={0,0,-1,1};    // up down left right
int dy[4]={1,-1,0,0};
struct Node
{
    int x,y;
    int step;
    friend bool operator <(Node x1,Node x2)    // 优先队列 按经过的门的次数排序
    {
        return x1.step>x2.step;
    }
}cur,now;
priority_queue<Node>q;

bool isok(int u,int v,int dd)           // 判断是否能走  要注意方向不同时 要判断不同额边
{
    int i,j;
    tstep=0;
    if(u+dx[dd]>maxx+1||u+dx[dd]<0||v+dy[dd]>maxy+1||v+dy[dd]<0||vis[u+dx[dd]][v+dy[dd]][dd]) return false;
    if(dd==0)
    {
        if(mpx[u][v+dy[dd]]==2) return false;
        else if(mpx[u][v+dy[dd]]) tstep=1;
    }
    else if(dd==1)
    {
        if(mpx[u][v]==2) return false;
        else if(mpx[u][v]) tstep=1;
    }
    else if(dd==2)
    {
        if(mpy[u][v]==2) return false;
        else if(mpy[u][v]) tstep=1;
    }
    else
    {
        if(mpy[u+dx[dd]][v]==2) return false;
        else if(mpy[u+dx[dd]][v]) tstep=1;
    }
    return true;
}
bool bfs()
{
    int i,xx,yy,cnt,flag1,fstep;
    ans=100000000;
    while(!q.empty()) q.pop();
    memset(vis,0,sizeof(vis));
    cur.x=0;
    cur.y=0;
    cur.step=0;
    flag1=0;
    q.push(cur);
    while(!q.empty())
    {
        now=q.top();
        xx=now.x;
        yy=now.y;
        cnt=now.step;
        if(xx==ex&&yy==ey)
        {
            flag1=1;
            fstep=cnt;
            if(ans>cnt) ans=cnt;
        }
        if(flag1)
        {
            if(cnt>fstep) return true;
        }
        for(i=0;i<4;i++)
        {
            if(isok(xx,yy,i))
            {
                cur.x=xx+dx[i];
                cur.y=yy+dy[i];
                cur.step=cnt+tstep;        // 经过门则加1 不经过则加0
                q.push(cur);
                vis[cur.x][cur.y][i]=1;
            }
        }
        q.pop();
    }
    return false;
}
int main()
{
    int i,j;
    int tx,ty,d,l;
    double t1,t2;
    char c;
    while(scanf("%d%d",&n,&m))
    {
        if(n==-1&&m==-1) break;
        memset(mpx,0,sizeof(mpx));
        memset(mpy,0,sizeof(mpy));
        maxx=maxy=-1;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d%d",&tx,&ty,&d,&l);
            if(d)
            {
                for(j=ty;j<ty+l;j++)
                {
                    mpy[tx][j]=2;                // 墙
                }
                if(maxy<ty+l-1) maxy=ty+l-1;
            }
            else
            {
                for(j=tx;j<tx+l;j++)
                {
                    mpx[j][ty]=2;
                }
                if(maxx<tx+l-1) maxx=tx+l-1;
            }
        }
        for(i=1;i<=m;i++)
        {
            scanf("%d%d%d",&tx,&ty,&d);
            if(d)
            {
                for(j=ty;j<ty+1;j++)
                {
                    mpy[tx][j]=1;                // 门
                }
            }
            else
            {
                for(j=tx;j<tx+1;j++)
                {
                    mpx[j][ty]=1;
                }
            }
        }
        scanf("%lf%lf",&t1,&t2);
        ex=int(t1);
        ey=int(t2);
        flag=ans=0;
        if(ex>maxx+1||ey>maxy+1||ex<0||ey<0) flag=1;
        if(flag||bfs()) printf("%d\n",ans);
        else  printf("-1\n");
    }
    return 0;
}