# Ignatius and the Princess I
**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15576 Accepted Submission(s): 4937
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
~~~
1.abduct：绑架，诱拐，使外展
2.如果结构体中有构造函数，用该结构体类型去定义变量时，必须用构造函数去初始化变量，否则编译通不过
3.代码：
~~~
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
using namespace std;
struct node
{
int x,y,t;
node(int a,int b,int c):x(a),y(b),t(c) {}//构造函数
void Set(int a,int b,int c)//设置函数
{
x=a;
y=b;
t=c;
}
bool operator <(const node &a)const//让优先队列的优先级：t小的，优先级高
{
return a.t<t;
}
};
struct node1
{
int x,y;
};
int n,m;
char mat[105][105];
bool vis[105][105];
node1 matt[105][105];//记录父亲坐标
int dir[4][2]= {1,0,0,1,-1,0,0,-1};
bool flag;
void bfs()
{
memset(vis,0,sizeof(vis));
priority_queue<node> Q;
node first=node(0,0,0);
Q.push(first);
vis[first.x][first.y]=1;
matt[0][0].x=0;
matt[0][0].y=0;
while(!Q.empty())
{
first=Q.top();
Q.pop();
if(first.x==n-1&&first.y==m-1)
{
flag=1;
printf("It takes %d seconds to reach the target position, let me show you the way.\n",first.t);
stack<node1> S;//把路径存储在栈里
node1 e;
e.x=n-1;
e.y=m-1;
S.push(e);
while(1)
{
e=S.top();
if(e.x==matt[e.x][e.y].x&&e.y==matt[e.x][e.y].y)
{
break;
}
node1 ee;
ee.x=matt[e.x][e.y].x;
ee.y=matt[e.x][e.y].y;
S.push(ee);
}
for(int i=1; i<=first.t; i++)//输出路径
{
printf("%ds:",i);
if(mat[S.top().x][S.top().y]!='.')
{
printf("FIGHT AT (%d,%d)\n",S.top().x,S.top().y);
mat[S.top().x][S.top().y]--;
if(mat[S.top().x][S.top().y]=='0')
{
mat[S.top().x][S.top().y]='.';
}
}
else
{
printf("(%d,%d)",S.top().x,S.top().y);
S.pop();
printf("->(%d,%d)\n",S.top().x,S.top().y);
}
}
printf("FINISH\n");
break;
}
node next=node(0,0,0);
for(int i=0; i<4; i++)
{
int tx=first.x+dir[i][0];
int ty=first.y+dir[i][1];
if(tx>=n||tx<0||ty>=m||ty<0||vis[tx][ty]==1)
{
continue;
}
else if(mat[tx][ty]=='X')
{
continue;
}
else if(mat[tx][ty]=='.')
{
next.Set(tx,ty,first.t+1);
matt[tx][ty].x=first.x;
matt[tx][ty].y=first.y;
vis[tx][ty]=1;
Q.push(next);
}
else
{
next.Set(tx,ty,first.t+1+mat[tx][ty]-'0');
vis[tx][ty]=1;
matt[tx][ty].x=first.x;
matt[tx][ty].y=first.y;
Q.push(next);
}
}
}
if(!flag)
printf("God please help our poor hero.\nFINISH\n");
}
int main()
{
while(scanf("%d%d",&n,&m)==2)
{
for(int i=0; i<n; i++)
{
scanf("%s",mat[i]);
}
flag=0;
bfs();
}
return 0;
}
~~~
- 前言
- The 12th Zhejiang Provincial Collegiate Programming Contest - D
- 用邻接表存储n个顶点m条弧的有向图
- hdu 5289 Assignment（给一个数组，求有多少个区间，满足区间内的最大值和最小值之差小于k）
- hdu 1358 Period（给定一个字符串，求有多少个前缀（包括自己本身），它是由k（k&gt;2,并且尽量大）个循环节组成的）
- hdu 1806 Frequent values(给定一个非降序数组，求任意区间内出现次数最多的数的次数)
- poj 3264 Balanced Lineup（查询区间最大值与最小值的差）
- HDU 1010 Tempter of the Bone（DFS+奇偶剪枝）
- HDU 1015 Safecracker(第一次用了搜索去遍历超时,第二次用for循环可以了,思路一样的)
- HDU 1016 Prime Ring Problem(DFS)
- HDU 1026 Ignatius and the Princess I(BFS+记录路径)
- HDU 1072 Nightmare（BFS）
- HDU 1237 简单计算器（后缀式+栈）