HDU 1026 Ignatius and the Princess I

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.XXXXX15 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)FINISHIt 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)FINISHGod please help our poor hero.FINISH

题意：给你一个图，从(0,0)出发，到达(n-1,m-1),X不能走，'.'可以走，但需要耗时1;数字是需要额外消耗时间，例如到达有个2这个点需要耗费3,问最少要多少时间;

思路：一看到这个题就首先想到了广搜加优先队列，看到输出格式时，还需要在广搜中输出路径，那么在广搜中怎样记录路径呢？我们在广搜中每个点最多只走一次。那么每个点的前驱有且只有一个，我们从终点开始找前驱，直道找到出发点的位置，即为反路径。那么该怎么找到前驱呢？因为在图中每次都只走到相邻的点，我们可以存下前驱的方向，就能找到前驱。(这个方向你需要自己去定)；

#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>using namespace std;int path[210][210];char map[210][210];int book[210][210];int dis[4][2]= {-1,0,1,0,0,-1,0,1}; //上下左右依次;int n,m;struct node{    int step;    int x,y;    friend bool operator <(node a,node b)    {        return a.step>b.step;    }};int bfs(){    priority_queue <node>Q;    node p,q;    p.x=p.y=0;    p.step=0;    book[0][0]=1;    Q.push(p);    while(Q.size())    {        p=Q.top();        Q.pop();        if(p.x==n-1&&p.y==m-1)            return p.step;        for(int i=0; i<4; i++)        {            int tx=p.x+dis[i][0];            int ty=p.y+dis[i][1];            if(tx<0||ty<0||tx>=n||ty>=m||book[tx][ty]||map[tx][ty]=='X')                continue;            book[tx][ty]=1;            q.x=tx;            q.y=ty;            path[tx][ty]=i;         //记录方向;            if(map[tx][ty]=='.')                q.step=p.step+1;            else                q.step=p.step+map[tx][ty]-'0'+1;            Q.push(q);        }    }    return -1;}int main(){    while(~scanf("%d %d",&n,&m))    {        for(int i=0; i<n; i++)            scanf("%s",map[i]);        memset(path,0,sizeof(path));        memset(book,0,sizeof(book));        int term=bfs();        if(term>-1)            printf("It takes %d seconds to reach the target position, let me show you the way.\n",term);        else        {            printf("God please help our poor hero.\nFINISH\n");            continue;        }        int S[100010][2];        int tx=n-1;        int ty=m-1,k=0;        while(1)        {            if(tx==0&&ty==0)                break;            int t=path[tx][ty];            if(map[tx][ty]=='.')            {                S[k][0]=tx;                S[k++][1]=ty;            }            else            {                int term=map[tx][ty]-'0';                while(term>=0)                {                    S[k][0]=tx;                    S[k++][1]=ty;                    term--;                }            }            if(t==0)                tx++;            else if(t==1)                tx--;            else if(t==2)                ty++;            else  if(t==3)                ty--;        }        S[k][0]=0;        S[k][1]=0;        for(int i=k-1,j=1; i>=0; i--,j++)        {            if(S[i][0]!=S[1+i][0]||S[i][1]!=S[1+i][1])                printf("%ds:(%d,%d)->(%d,%d)\n",j,S[i+1][0],S[i+1][1],S[i][0],S[i][1]);            else printf("%ds:FIGHT AT (%d,%d)\n",j,S[i][0],S[i][1]);        }        printf("FINISH\n");    }    return 0;}