Ignatius and the Princess I（bfs+记录路径）hdu 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.
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
算是做过的第一道正式记录路径的题目，还是纪念一下吧，虽然说，，，没有完全正确，，，
先贴上我的代码（不正确！！！）
//入口和出口也可能有怪兽，记录路径啊~~记录路径~~
//bfs+优先队列
#include
#include
#include
using namespace std;
struct node{
int x,y;
int times;
};
bool operator<(const node &a,const node &b)
{
return a.times>b.times;
}
int n,m,step;
char map[105][105];
int vis[105][105];
int fx[105],fy[105],k;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
int judge(node p)
{
if(p.x<0 || p.x==n || p.y<0 || p.y==m)
return false;
if(map[p.x][p.y]=='X' || vis[p.x][p.y])
return false;
return true;
}
void bfs()
{
priority_queueQ;
node q,p;
q.x=q.y=0;
q.times=0;
if(map[q.x][q.y]!='.' && map[q.x][q.y]!='X')
q.times+=map[q.x][q.y]-'0';
vis[q.x][q.y]=1;
Q.push(q);
while(!Q.empty())
{
q=Q.top();
Q.pop();
fx[k]=q.x,fy[k++]=q.y;
//printf("%d %d %d\n",q.x,q.y,q.times);
if(q.x==n-1 && q.y==m-1)
{
step=q.times;
return ;
}
for(int i=0;i<4;i++)
{
p=q;
p.x+=dx[i];
p.y+=dy[i];
p.times++;
if(!judge(p))
continue;
if(map[p.x][p.y]!='.' && map[p.x][p.y]!='X')
p.times+=map[p.x][p.y]-'0';
vis[p.x][p.y]=1;
Q.push(p);
}
}
}
void printpath()
{
printf("It takes %d seconds to reach the target position, let me show you the way.\n",step);
int t=1;
for(int i=0;i(%d,%d)\n",t++,fx[i],fy[i],fx[i+1],fy[i+1]);
if(map[fx[i+1]][fy[i+1]]!='.' && map[fx[i+1]][fy[i+1]]!='X')
{
int tt=map[fx[i+1]][fy[i+1]]-'0';
while(tt>0)
{
printf("%ds:FIGHT AT (%d,%d)\n",t++,fx[i+1],fy[i+1]);
tt--;
}
}
}
}
int main()
{
while(scanf("%d %d",&n,&m)!=EOF)
{
memset(map,0,sizeof(map));
memset(vis,0,sizeof(vis));
memset(fx,0,sizeof(fx));
memset(fy,0,sizeof(fy));
for(int i=0;i 参考了大神的代码（大神说是简单的bfs，，，_QAQ_）之后，我很认真的分析了路径的保存问题
注意：
1、因为打怪兽要消耗时间，所以要用优先队列记录状态
2、需要输出路径，所以用flag[ i ] [ j ]记录朝向
关于路径问题，我通过画图发现，每个可以走的点都会被遍历到并且只能被遍历一次，但是从
起点到终点只有一条完整的路径。看下图，画叉的是障碍，三角中的数字是怪物的血量，其他每个
圆里的数字是走到该点所用的时间。我们发现，按照箭头的方向，我们只能找到一条从起点通往终
点的路，那么我们可以用一个二维数组flag记录下箭头的指向，最后从终点递归到起点，返回时输
出，就可以得到路径
5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX.
附上AC代码：
#include
#include
#include
using namespace std;
struct node
{
int x,y;
int step;
};
bool operator<(node a,node b)
{
return a.step>b.step;
}
int map[105][105]; //地图
int flag[105][105];//flag[i][j]记录是怎么从上一个位置走到(i,j)的, 1代表(向下),2代表(向上),3代表(向左),4代表(向右)
int cnt[105][105];//记录怪的血量
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
int n,m,tim;
int check(int x,int y) //判断(x,y)是否可走
{
if(x<0 || y<0 || x>=n || y>=m)
return 1;
if(map[x][y] == -1)
return 1;
return 0;
}
int bfs()
{
int i;
priority_queue Q;
node a,next;
a.x = 0;
a.y = 0;
a.step = 0;
map[0][0] = -1;
Q.push(a);
while(!Q.empty())
{
a = Q.top();
Q.pop();
if(a.x == n-1 && a.y == m-1)
return a.step;
for(i = 0; i<4; i++)
{
next=a;
next.x+=dx[i];
next.y+=dy[i];
if(check(next.x,next.y))
continue;
next.step =a.step+map[next.x][next.y]+1;//记录时间
map[next.x][next.y]=-1;//标记为走过
flag[next.x][next.y]=i+1;//记录朝向
Q.push(next);
}
}
return 0;
}
void print(int x,int y) //递归输出路径
{
int n_x,n_y;
if(!flag[x][y])//回到起点，返回
return;
n_x = x - dx[flag[x][y]-1];
n_y = y - dy[flag[x][y]-1];
print(n_x,n_y);
printf("%ds:(%d,%d)->(%d,%d)\n",tim++,n_x,n_y,x,y);
while(cnt[x][y]--)
{
printf("%ds:FIGHT AT (%d,%d)\n",tim++,x,y);
}
}
int main()
{
int i,j;
while(~scanf("%d%d",&n,&m))
{
memset(map,0,sizeof(map));
memset(flag,0,sizeof(flag));
memset(cnt,0,sizeof(cnt));
char s[105];
for(i = 0; i