题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1026

Ignatius and the Princess I

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

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

网上说用stack输出路径

可是stack还不会用，参照了大神的思路，只要倒着搜索就可以不用栈了，用二维数组记录每一步的前驱再往回打印就可以输出路径了；

本题用到了优先队列，时间小的先出队，对优先队列有疑惑的同学请移步 ：STL优先队列详解

【源代码】

#include <stdio.h>
#include <cstring>
#include <queue>
using namespace std;
#define maxn 110
char map[110][110];
bool vis[maxn][maxn];
struct point
{
    int x,y;int times;
    friend bool operator <(point a,point b)
    {
        return a.times>b.times;  //重载小于号，使得时间小的先出队列；
    }
};
struct Pre{
    int px,py;
}pre[maxn][maxn];
int r,c;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
bool inborder(point a)
{
    if(a.x<0||a.y<0||a.x>r-1||a.y>c-1)
        return false;
    return true;
}
void BFS(int x, int y)
{
    pre[x][y].px=-1;
    vis[x][y]=1;
    point st;
    st.x=x,st.y=y;
    if(map[st.x][st.y]!='.')
        st.times=(map[st.x][st.y]-'0');
    else
        st.times=0;
    priority_queue<point>Q;
    while(!Q.empty())
        Q.pop();
    Q.push(st);
    while(!Q.empty())
    {
        point now=Q.top();
        Q.pop();
        if(now.x==0&&now.y==0)
        {
            int key=1;
            printf("It takes %d seconds to reach the target position, let me show you the way.\n",now.times);
            while(pre[now.x][now.y].px!=-1)//循环找前驱，直到终点
            {
                int tx=pre[now.x][now.y].px;
                int ty=pre[now.x][now.y].py;
                if(map[tx][ty]=='.')
                {
                    printf("%ds:(%d,%d)->(%d,%d)\n",key++,now.x,now.y,tx,ty);
                }
                else//打怪耗时
                {
                    printf("%ds:(%d,%d)->(%d,%d)\n",key++,now.x,now.y,tx,ty);
                    int tt=map[tx][ty]-'0';
                    while(tt--)
                        printf("%ds:FIGHT AT (%d,%d)\n",key++,tx,ty);
                }
                now.x=tx;//将当前前驱作为起始
                now.y=ty;
            }
            //printf("%ds:(%d,%d)->(%d,%d)\n",key++,now.x,now.y,r-1,c-1);
            return ;
        }
        else
        {
            point next;
            for(int i=0;i<4;i++)
            {
                next.x=now.x+dx[i];
                next.y=now.y+dy[i];
                next.times=now.times;
                if(map[next.x][next.y]=='X'||vis[next.x][next.y]||!inborder(next))
                    continue;
                else
                {
                    vis[next.x][next.y]=1;
                    if(map[next.x][next.y]=='.')
                    {
                        next.times+=1;
                    }
                    else
                    {
                        next.times+=(map[next.x][next.y]-'0'+1);//除了打怪耗时，还有走到该坐标的耗时故+1；
                    }
                    pre[next.x][next.y].px=now.x;//记录前驱
                    pre[next.x][next.y].py=now.y;
                    Q.push(next);
                }
            }
        }
    }
    printf("God please help our poor hero.\n");
}
void init()
{
    memset(vis,0,sizeof(vis));
    memset(map,'\0',sizeof(map));
    memset(pre,0,sizeof(pre));
}
int main()
{
    //int r,c;
    while(~scanf("%d%d",&r,&c))
    {
        init();
        for(int i=0;i<r;i++)
            scanf("%s",map[i]);
        BFS(r-1,c-1);//从右下角搜索
        puts("FINISH");
    }
    return 0;
}