BFS最短路径----迷宫问题

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

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

#include
#include
#define MAX 1000000
using namespace std;
struct Path{
    int time;       //表示这点需要的时间
    int x,y;        //用来指示父节点map[i][j].x  表示这个点的父节点x值，y同理, niubi
    int ttime;      //map[i][j].ttime=表示总共的时间，totaltime
}map[105][105];

struct Point{
    int x,y;         //用来指示地图的位置
}p1,p2;

int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int row,col;
void bfs();
int main(){
    char ch;
    cin>>row>>col;
    getchar();
    for(int i=0;i<row;i++){
        for(int j=0;j<col;j++){
            cin>>ch;
            map[i][j].ttime=MAX;                    //找到最小值，这里设置为最大
            if(ch=='.'){map[i][j].time=1;}
            else if(ch=='X'){map[i][j].time=-1;}
            else map[i][j].time=ch-'0'+1;
        }
        getchar();
    }
    bfs();
    if(map[0][0].ttime==MAX){printf("God please help our poor hero.\n");}      //说明没有走回到原点
    else{           //又要写出路径，也要写成打怪兽的地点
        printf("It takes %d seconds to reach the target position, let me show you the way.\n",map[0][0].ttime);
        int i=0,j=0,k=0,h=0;
        while(i<map[0][0].ttime){
            if(map[j][k].time>1){
                while(--map[j][k].time){
                    i++;
                    printf("%ds:FIGHT AT (%d,%d)\n",i,j,k);
                }
            }else{
                i++;
                //printf("j=%d,k=%d\n",j,k);
                printf("%ds:(%d,%d)->(%d,%d)\n",i,j,k,map[j][k].x,map[j][k].y);
                h=map[j][k].x;
                k=map[j][k].y;
                j=h;
            }
        }
        printf("FINISH\n");
    }
    return 0;
}

void bfs(){
    queue<Point>q;
    map[0][0].time=0;
    p1.x=row-1;
    p1.y=col-1;
    map[row-1][col-1].ttime=map[row-1][col-1].time;
    q.push(p1);
    while(!q.empty()){
        p1=q.front();
        q.pop();
        if(p1.x==0&&p1.y==0)continue;       //回到原点，不进入计算，也就是控制不再次反向运算修改值
        for(int k=0;k<4;k++){
            p2.x=p1.x+dir[k][0];
            p2.y=p1.y+dir[k][1];
            if(p2.x<0||p2.x>row-1||p2.y<0||p2.y>col-1||map[p2.x][p2.y].time==-1){continue;}//如果出墙，或者是一个trap的话，那么后面的就不要计算的了
            if(map[p1.x][p1.y].ttime+map[p2.x][p2.y].time<map[p2.x][p2.y].ttime){
                map[p2.x][p2.y].ttime=map[p1.x][p1.y].ttime+map[p2.x][p2.y].time;
                map[p2.x][p2.y].x=p1.x;
                map[p2.x][p2.y].y=p1.y;
                q.push(p2);
            }
        }
    }
}