hdu 1026 Ignatius and the Princess I （bfs记录路径 两种记录路径方法模拟优先队列和优先队列两种做法）

Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9433    Accepted Submission(s): 2784

Special Judge

Problem Description

The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-top corner is (0,0) and right-bottom corner is (N-1,M-1). Ignatius enters at (0,0), and the door to feng5166's room is at (N-1,M-1), that is our target. There are some monsters in the castle, if Ignatius meet them, he has to kill them. Here is some rules:

1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.

Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position.

 

Input

The input contains several test cases. Each test case starts with a line contains two numbers N and M(2<=N<=100,2<=M<=100) which indicate the size of the labyrinth. Then a N*M two-dimensional array follows, which describe the whole labyrinth. The input is terminated by the end of file. More details in the Sample Input.

 

Output

For each test case, you should output "God please help our poor hero." if Ignatius can't reach the target position, or you should output "It takes n seconds to reach the target position, let me show you the way."(n is the minimum seconds), and tell our hero the whole path. Output a line contains "FINISH" after each test case. If there are more than one path, any one is OK in this problem. More details in the Sample Output.

 

Sample Input

   
   
   
   

    
    
    
    5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX.
5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX1
5 6
.XX...
..XX1.
2...X.
...XX.
XXXXX.
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    It takes 13 seconds to reach the target position, let me show you the way.
1s:(0,0)->(1,0)
2s:(1,0)->(1,1)
3s:(1,1)->(2,1)
4s:(2,1)->(2,2)
5s:(2,2)->(2,3)
6s:(2,3)->(1,3)
7s:(1,3)->(1,4)
8s:FIGHT AT (1,4)
9s:FIGHT AT (1,4)
10s:(1,4)->(1,5)
11s:(1,5)->(2,5)
12s:(2,5)->(3,5)
13s:(3,5)->(4,5)
FINISH
It takes 14 seconds to reach the target position, let me show you the way.
1s:(0,0)->(1,0)
2s:(1,0)->(1,1)
3s:(1,1)->(2,1)
4s:(2,1)->(2,2)
5s:(2,2)->(2,3)
6s:(2,3)->(1,3)
7s:(1,3)->(1,4)
8s:FIGHT AT (1,4)
9s:FIGHT AT (1,4)
10s:(1,4)->(1,5)
11s:(1,5)->(2,5)
12s:(2,5)->(3,5)
13s:(3,5)->(4,5)
14s:FIGHT AT (4,5)
FINISH
God please help our poor hero.
FINISH
   
   
   
   

题目大意：

    模拟一个人走迷宫，遇到怪兽要和他决斗，决斗时间为那个格子的数字。输出(0,0)->(n-1,m-1)的最短路并输出路径。

感想：

    题目并不是很难，本来可以一个小时之内过的，结果坑了我一下午。主要是我把存答案的数组开小了 。还有有两种存路径的方法  见下面的代码

代码1： 自己没用STL的代码    用结构存上一个点的坐标

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
const int INF=0x3f3f3f3f;
#define maxn 105
using namespace std;

int n,m,cxx;
int sx,sy,ex,ey;
int mp[maxn][maxn];   // 地图
int vis[maxn][maxn];  // 地图标记
int sta[maxn*maxn];   // 栈
int vissta[maxn*maxn];  // 标记栈是否访问
int dx[4]= {-1,1,0,0};  //　方向数组
int dy[4]= {0,0,-1,1};
char s[maxn],ans[1000005][40];　// 答案数组  注意一定不能吝啬内存 不然你会被坑的很惨
struct Tnode
{
    int cnt;          // 步数
    int prex,prey;    // 记录前一个点
} node[maxn][maxn];

bool bfs()
{
    int i,x,y,nx,ny,tempx,tempy;
    int mi,k;
    int head=0,tail=-1;
    sta[++tail]=(sx-1)*m+sy;       // 将行列对应值进栈
    memset(vis,0,sizeof(vis));
    memset(vissta,0,sizeof(vissta));
    vis[sx][sy]=1;
    node[sx][sy].cnt=0;
    node[sx][sy].prex=-1;
    node[sx][sy].prey=-1;
    while(head<=tail)
    {
        mi=100000000;
        for(i=head; i<=tail; i++)    // 模拟优先队列 找到栈中步数最小的
        {
            tempy=sta[i]%m;
            if(tempy==0) tempy=m;
            tempx=(sta[i]-tempy)/m+1;
            if(node[tempx][tempy].cnt<mi&&!vissta[i])
            {
                mi=node[tempx][tempy].cnt;
                k=i;
            }
        }
        y=sta[k]%m;            // 计算栈对应的行列
        if(y==0) y=m;
        x=(sta[k]-y)/m+1;
        vissta[k]=1;           // 标记栈已经访问
        while(vissta[head]) head++;   // 移动栈头
        if(x==ex&&y==ey)
        {
            return true;
        }
        for(i=0; i<4; i++)     // bfs搜索进栈
        {
            nx=x+dx[i];
            ny=y+dy[i];
            if(mp[nx][ny]<10&&!vis[nx][ny])
            {
                sta[++tail]=(nx-1)*m+ny;
                vis[nx][ny]=1;
                node[nx][ny].prex=x;
                node[nx][ny].prey=y;
                node[nx][ny].cnt=node[x][y].cnt+mp[nx][ny]+1;
            }
        }
    }
    return false;
}
void solve(int t)    // 找路径 往上推 存答案
{
    int i,j,temp;
    int su,sv,eu=ex,ev=ey;
    cxx=t+1;
    while(1)
    {
        su=node[eu][ev].prex;
        sv=node[eu][ev].prey;
        if(su==-1) break;
        if(mp[eu][ev]==0)
        {
            cxx--;
            sprintf(ans[cxx],"%ds:(%d,%d)->(%d,%d)",cxx,su-1,sv-1,eu-1,ev-1);
        }
        else
        {
            temp=mp[eu][ev];
            while(temp--)
            {
                cxx--;
                sprintf(ans[cxx],"%ds:FIGHT AT (%d,%d)",cxx,eu-1,ev-1);
            }
            cxx--;
            sprintf(ans[cxx],"%ds:(%d,%d)->(%d,%d)",cxx,su-1,sv-1,eu-1,ev-1);
        }
        eu=su;
        ev=sv;
    }
}
int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        getchar();
        memset(mp,0x3f,sizeof(mp));  // 初始化地图都不可以走
        for(i=1; i<=n; i++)          // 我将题图整体往斜下方移了一个单位 就能自动判断出界了
        {
            scanf("%s",s);
            for(j=1; j<=m; j++)
            {
                if(s[j-1]=='.') mp[i][j]=0;
                else if(s[j-1]>='1'&&s[j-1]<='9') mp[i][j]=s[j-1]-'0';
            }
        }
        sx=1;
        sy=1;
        ex=n;
        ey=m;
        if(bfs())
        {
            printf("It takes %d seconds to reach the target position, let me show you the way.\n",node[ex][ey].cnt);
            solve(node[ex][ey].cnt);
            for(i=1;i<=node[ex][ey].cnt;i++)
                printf("%s\n",ans[i]);
        }
        else printf("God please help our poor hero.\n");
        printf("FINISH\n");
    }
    return 0;
}

代码2：用优先队列的代码   用dir[ ][ ]数组存上一个点的方向

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#define maxn 105
using namespace std;

const int INF=0x3f3f3f3f;
int n,m,cxx;
int mp[maxn][maxn];      // 地图
int vis[maxn][maxn];     // 标记数组
int dir[maxn][maxn];     // 记录方向数组
int dx[4]= {-1,1,0,0};
int dy[4]= {0,0,-1,1};
char s[maxn],ans[1000005][40];
struct Tnode
{
    int cnt;
    int curx,cury;
    friend bool operator < (Tnode a,Tnode b)
    {
        return a.cnt>b.cnt;
    }
} t1,now;
priority_queue <Tnode> q;

bool bfs()
{
    int i,x,y,nx,ny;
    while(!q.empty()) q.pop();
    t1.cnt=0;
    t1.curx=1;
    t1.cury=1;
    q.push(t1);
    memset(vis,0,sizeof(vis));
    dir[1][1]=-1;
    vis[1][1]=1;
    while(!q.empty())
    {
        now=q.top();     // 用优先队列可以保证每次出栈的都是步数最小的
        y=now.cury;
        x=now.curx;
        q.pop();
        if(x==n&&y==m)
        {
            cxx=now.cnt;
            return true;
        }
        for(i=0; i<4; i++)
        {
            nx=x+dx[i];
            ny=y+dy[i];
            if(mp[nx][ny]<INF&&!vis[nx][ny])
            {
                vis[nx][ny]=1;
                dir[nx][ny]=i;
                t1.cnt=now.cnt+mp[nx][ny]+1;
                t1.curx=nx;
                t1.cury=ny;
                q.push(t1);
            }
        }
    }
    return false;
}
void solve()
{
    int i,j,temp,t;
    int su,sv,eu=n,ev=m;
    t=cxx+1;
    while(1)
    {
        if(t<1) break;
        su=eu-dx[dir[eu][ev]];    // 通过方向找到上一个点的坐标
        sv=ev-dy[dir[eu][ev]];
        if(mp[eu][ev]>0)
        {
            temp=mp[eu][ev];
            while(temp--)
            {
                t--;
                sprintf(ans[t],"%ds:FIGHT AT (%d,%d)",t,eu-1,ev-1);
            }
        }
        t--;
        sprintf(ans[t],"%ds:(%d,%d)->(%d,%d)",t,su-1,sv-1,eu-1,ev-1);
        eu=su;
        ev=sv;
    }
}
int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        getchar();
        memset(mp,0x3f,sizeof(mp));
        for(i=1; i<=n; i++)
        {
            scanf("%s",s);
            for(j=1; j<=m; j++)
            {
                if(s[j-1]=='.') mp[i][j]=0;
                else if(s[j-1]>='1'&&s[j-1]<='9') mp[i][j]=s[j-1]-'0';
            }
        }
        if(bfs())
        {
            printf("It takes %d seconds to reach the target position, let me show you the way.\n",cxx);
            solve();
            for(i=1; i<=cxx; i++)
                printf("%s\n",ans[i]);
        }
        else printf("God please help our poor hero.\n");
        printf("FINISH\n");
    }
    return 0;
}