HDU1026

Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10225    Accepted Submission(s): 3095
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
   
   
   
   

Author

Ignatius.L

 

   题意:题意大致就是找从最左上角A[0,0]走到最右下角A[n-1][m-1]的最短带权路径.求最短路径一般基于BFS,本来打算用Dijkstra算法做.但是,这道题的输出要求很苛刻,貌似这是HDU的风格,输出要求一般比较苛刻.但是思路一样是基于BFS. 我们把图的每一个结点保存当前坐标(x,y),当前结点后继坐标(nextX,nextY),当前字符cc,当前最短权重time.从最末终的结点开始宽度优先搜索,一层一层往外扩,直到搜索完成,就可以生成一棵倒序的宽度优先搜索树,A[n-1][m-1]结点为树根,而A[0][0]为树叶,只要找从A[0][0]到A[n-1][m-1]的路径即为带树最短路径.

BFS思路明了.但是细节决定成败,真是WA了好多次.

#include<iostream>
#include<queue>
#include<limits.h>
using namespace std;
int n, m;
struct node
{
    char cc;
    int x;//当前坐标
    int y;
    int nextX;//后继坐标
    int nextY;
    int time;
};
int dir[4][2]={{0,1},{0,-1},{-1,0},{1,0}}; //4方向
node map[105][105];

void outPut()//吭爹的output
{
    int x=0,y=0,a,b,time=1,i;
    if(map[0][0].time!= INT_MAX)
    {
        printf("It takes %d seconds to reach the target position, let me show you the way.\n",map[0][0].time);
        while(x!= n-1|| y!=m-1)
        {
            a=map[x][y].nextX; 
            b=map[x][y].nextY;
            if(map[x][y].cc!='.')
                for(i=0;i<map[x][y].cc - '0';i++)
                    printf("%ds:FIGHT AT (%d,%d)\n",time++,x,y);
            printf("%ds:(%d,%d)->(%d,%d)\n",time++,x,y,a,b);
            x=a; y=b;
        }
        if(map[x][y].cc!='.')
        {
            for(i=0;i<map[x][y].cc - '0';i++)
                printf("%ds:FIGHT AT (%d,%d)\n",time++,x,y);
        }
        printf("FINISH\n");
    }
    else
    {
        printf("God please help our poor hero.\nFINISH\n");
    }
}
void BFS()
{
    map[n-1][m-1].time=0;
    queue<node> Q;
    node qhead;
    int newPathLen=0;
    if( map[n-1][m-1].cc >='1' && map[n-1][m-1].cc <='9')//当前结点符号是数值,从终于开始搜索
        map[n-1][m-1].time=map[n-1][m-1].cc-'0';
    Q.push(map[n-1][m-1]);
    while(!Q.empty())
    {
        qhead=Q.front();
        Q.pop();
        for(int i=0;i<4;++i)//4 direction
        {
            int tx=qhead.x+dir[i][0],ty=qhead.y+dir[i][1];//(tx,ty)是新坐标
            if( tx < 0 || ty <0 ||tx>=n || ty >= m || map[tx][ty].cc =='X' )//新坐标在图外或者是X
                continue;
            if( map[tx][ty].cc != '.')//如果是数值,则权值+1,进一步要+1
                newPathLen =map[tx][ty].cc-'0' + qhead.time ;//如果不是数值,权值不变
            else newPathLen = qhead.time ;//如果新的路径比原路径更短,更新之
            if( newPathLen+1 < map[tx][ty].time)
            {
                map[tx][ty].time=newPathLen+1;
                map[tx][ty].nextX = qhead.x;
                map[tx][ty].nextY = qhead.y;
                Q.push(map[tx][ty]);
            }
        }
    }
    outPut();
}
int main()
{
    while(cin>>n>>m)
    {
        for(int i=0;i<n;++i)
        {
            for(int j=0;j<m;++j)
            {
                cin>>map[i][j].cc;
                map[i][j].time=INT_MAX;
                map[i][j].x=i;
                map[i][j].y=j;
                map[i][j].nextX=-1;
                map[i][j].nextY=-1;
            }
        }
        BFS();
    }
    
    return 0;
}