HDU-1026 Ignatius and the Princess I(BFS) 带路径的广搜
此题需要时间更少,控制时间很要,这个题目要多多看,
Ignatius and the Princess I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9910 Accepted Submission(s): 2959 Special Judge
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.
1s:(0,0)->(1,0)
It takes 14 seconds to reach the target position, let me show you the way.
1
God please help our poor hero.
#include <stdio.h>
#define MAX_ 100000;
char map[][];
typedef struct {
int x ,y ,time ,front;
}point;
int dir[][] = {-,,,,,,,-},mintime[][];
int n , m ,I ,J;
point arr[];
void bfs ()
{
while (I != J)//队列不为空
{
point head = arr[I++];
for (int i = ;i <= ;i++)
{
int x = head.x + dir[i][] ,y = head.y + dir[i][];
if (x >= && y >= && x < m && y < n && map[x][y] != 'X')
{
point k;
//k.x = x ,k.y = y;
k.time = head.time + ;
if (map[x][y] != '.')
k.time += map[x][y] - '';
if(k.time < mintime[x][y])//若到达minttime【x】【y】时间小于其他点到此点时间,此点入队
{
mintime[x][y] = k.time;
k.x = x ,k.y = y;
k.front = I - ;
//printf("%d---%d---%d---%d\n",k.x ,k.y ,k.time ,k.front);
arr[J++] = k;
}
} }
}
}
int main()
{
while (scanf("%d %d",&m,&n) != EOF)
{
for (int i = ;i < m ;i++)
for (int j = ;j < n ;j++)
{
scanf(" %c",&map[i][j]);
mintime[i][j] = MAX_;
}
//队列第一个元素赋值
arr[].x = m - ,arr[].y = n - ,arr[].time = ,arr[].front = - ,mintime[m-][n-] = ;
if (map[m-][n-] != 'X' && map[m-][n-] != '.') mintime[m - ][n - ] = arr[].time = map[m-][n-] - '';
I = ,J = ;
int time = ;
//执行搜索
bfs ();
//不能到达
if (mintime[][] == )
printf ("God please help our poor hero.\n");
else
{
printf ("It takes %d seconds to reach the target position, let me show you the way.\n",mintime[][]);
point s = arr[I-];
while (s.x != || s.y !=)
s = arr[--I];
int p ,time = ,x ,y;
//打印路径
while (s.front >= )
{
p = s.front;
x = arr[p].x ,y = arr[p].y;
printf ("%ds:(%d,%d)->(%d,%d)\n",time++,s.x,s.y,x,y);
if (map[x][y] != 'X' && map[x][y] != '.')
for(int i = ;i <= map[x][y]-'';i++)
printf("%ds:FIGHT AT (%d,%d)\n",time++,x,y);
s = arr[p];
}
}
printf ("FINISH\n");
}
return ;
}
方法2:
#include <stdio.h>
#define MAX_ 100000;
char map[][];
typedef struct {
int x ,y ,time ,front;
}point;
int dir[][] = {-,,,,,,,-},mintime[][];
int n , m ,I ,J;
point arr[];
void bfs ()
{
while (I != J)//队列不为空
{
point head = arr[I++];
for (int i = ;i <= ;i++)
{
int x = head.x + dir[i][] ,y = head.y + dir[i][];
if (x >= && y >= && x < m && y < n && map[x][y] != 'X')
{
point k;
//k.x = x ,k.y = y;
k.time = head.time + ;
if (map[x][y] != '.')
k.time += map[x][y] - '';
if(k.time < mintime[x][y])//若到达minttime【x】【y】时间小于其他点到此点时间,此点入队
{
mintime[x][y] = k.time;
k.x = x ,k.y = y;
k.front = I - ;
//printf("%d---%d---%d---%d\n",k.x ,k.y ,k.time ,k.front);
arr[J++] = k;
}
} }
}
}
int main()
{
while (scanf("%d %d",&m,&n) != EOF)
{
for (int i = ;i < m ;i++)
for (int j = ;j < n ;j++)
{
scanf(" %c",&map[i][j]);
mintime[i][j] = MAX_;
}
//队列第一个元素赋值
arr[].x = m - ,arr[].y = n - ,arr[].time = ,arr[].front = - ,mintime[m-][n-] = ;
if (map[m-][n-] != 'X' && map[m-][n-] != '.') mintime[m - ][n - ] = arr[].time = map[m-][n-] - '';
I = ,J = ;
int time = ;
//执行搜索
bfs ();
//不能到达
if (mintime[][] == )
printf ("God please help our poor hero.\n");
else
{
printf ("It takes %d seconds to reach the target position, let me show you the way.\n",mintime[][]);
point s = arr[I-];
while (s.x != || s.y !=)
s = arr[--I];
int p ,time = ,x ,y;
//打印路径
while (s.front >= )
{
p = s.front;
x = arr[p].x ,y = arr[p].y;
printf ("%ds:(%d,%d)->(%d,%d)\n",time++,s.x,s.y,x,y);
if (map[x][y] != 'X' && map[x][y] != '.')
for(int i = ;i <= map[x][y]-'';i++)
printf("%ds:FIGHT AT (%d,%d)\n",time++,x,y);
s = arr[p];
}
}
printf ("FINISH\n");
}
return ;
}
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