HDU【1026】Ignatius and the Princess I

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Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15492 Accepted Submission(s): 4904
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.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

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <stack>
using namespace std;

const int maxn = 105;
char s[maxn][maxn];
int n,m,x,y,str[4][2] = {{1,0},{-1,0},{0,-1},{0,1}}; //向4个方向移动的2维数组

struct Node //定义一个储存当前结点x,y坐标的结构体
{
int xx,yy,time;
bool operator < (const Node no) const //重载小于运算符，使优先队列中的所有的点按点的时间从小到大排列
{
return time > no.time;
}
}node,pre;
Node p[maxn][maxn]; //存储上一步的位置

int BFS()
{
priority_queue <Node> q;
node.xx = 0;
node.yy = 0;
node.time = 0;
q.push(node);
s[0][0] = 'X';
while(!q.empty()) //通过入队出队实现广搜
{
node = q.top();
q.pop();
for(int i = 0 ; i < 4; i++)
{
x = node.xx + str[i][0];
y = node.yy + str[i][1];
if(x >= 0 && x < n && y >= 0 && y < m && s[x][y] != 'X')//如果该结点，没有超过边界同时没有被标记
{
if(x == n-1 && y == m-1) //考虑2种情况
{
if(s[x][y] == '.')
{
p[x][y].xx = node.xx;
p[x][y].yy = node.yy;
p[x][y].time = 1;
printf("It takes %d seconds to reach the target position, let me show you the way.\n",node.time + p[x][y].time);
}
else
{
p[x][y].xx = node.xx;
p[x][y].yy = node.yy;
p[x][y].time = 1 + s[x][y] - '0';
printf("It takes %d seconds to reach the target position, let me show you the way.\n",node.time + p[x][y].time);
}
return -1;
}
if(s[x][y] == '.')
{
p[x][y].xx = node.xx;
p[x][y].yy = node.yy;
p[x][y].time = 1;
pre.time = node.time + p[x][y].time;
}
else
{
p[x][y].xx = node.xx;
p[x][y].yy = node.yy;
p[x][y].time = 1 + s[x][y] - '0';
pre.time = node.time + p[x][y].time;
}
pre.xx = x;
pre.yy = y;
q.push(pre); //入队
s[x][y] = 'X'; //每一次经过的点均将该结点标记为'X';
}
}
}
}

int main()
{
while(scanf("%d%d",&n,&m) != EOF)
{
for(int i = 0 ; i < n; i++)
scanf("%s",s[i]);
if(BFS() != -1) cout << "God please help our poor hero." << endl;//无法到达目标地点
else
{
stack <Node> s; //通过栈来查询从起点到终点的路径
int x = n - 1;
int y = m - 1;
while(p[x][y].xx != 0 || p[x][y].yy != 0)
{
s.push(p[x][y]);
int x1 = x;
int y1 = y;
x = p[x1][y1].xx;
y = p[x1][y1].yy;
}
s.push(p[x][y]); //这时的p[x][y].x==0&&p[x][y].y==0也需要加入队列
int t = 0;
Node node1,node2;
while(s.size() >= 2) //因为里面进行了2次取栈首位置元素，所以需要保证栈里面至少有2个元素
{
node1 = s.top();
s.pop();
node2 = s.top();
if(node1.time == 1)
printf("%ds:(%d,%d)->(%d,%d)\n",++t,node1.xx,node1.yy,node2.xx,node2.yy); //不需要打怪
else
{
printf("%ds:(%d,%d)->(%d,%d)\n",++t,node1.xx,node1.yy,node2.xx,node2.yy); // 需要打怪
while((node1.time--) != 1)
printf("%ds:FIGHT AT (%d,%d)\n",++t,node2.xx,node2.yy);
}
}
if(node2.time == 1) //到达(n-1，m-1)的路径同样需要处理
printf("%ds:(%d,%d)->(%d,%d)\n",++t,node2.xx,node2.yy,n - 1,m - 1);
else
{
printf("%ds:(%d,%d)->(%d,%d)\n",++t,node2.xx,node2.yy,n - 1,m - 1);
while((node2.time--) != 1)
printf("%ds:FIGHT AT (%d,%d)\n",++t,n - 1,m - 1);
}
}
cout << "FINISH" << endl;
}
return 0;
}

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