ACM-BFS之Ignatius and the Princess I ——hdu1026
Ignatius and the Princess I
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. 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
经典的广搜题，用到了优先队列和打表存储，要注意优先队列的弹出队首元素是 .top()，
再注意一下开的数据大小问题，慢慢来，就可以AC
// 数据空间都要开大一点哟，刚开始我开的30，30结果一直WA
#include
#include
using namespace std;
struct Coordinate
{
int x,y,step;
// 优先队列
friend bool operator<(Coordinate c1,Coordinate c2)
{
return c2.step=m || y>=n) return 1;
if(visit[x][y]==1 || map[x][y]=='X') return 1;
return 0;
}
// 输出
void show(int a,int b,int t)
{
if(t==1) printf("%ds:(%d,%d)->(%d,%d)\n",t,path[a][b].x,path[a][b].y,a,b);
else if(path[a][b].step==1)
{
show(path[a][b].x,path[a][b].y,t-1);
printf("%ds:(%d,%d)->(%d,%d)\n",t,path[a][b].x,path[a][b].y,a,b);
}
else
{
--path[a][b].step;
show(a,b,t-1);
printf("%ds:FIGHT AT (%d,%d)\n",t,a,b);
}
}
// BFS 广搜
void bfs(void)
{
memset(visit,0,sizeof(visit));
priority_queue cd;
Coordinate p,next;
int i;
p.x=0,p.y=0,p.step=0;
visit[p.x][p.y]=1;
cd.push(p);
while(!cd.empty())
{
p=cd.top(); // 优先队列，所以这块不是cd.front()了
cd.pop();
if(p.x==m-1 && p.y==n-1)
{
ans=p.step;
isfind=1;
return;
}
for(i=0;i<4;++i)
{
next.x=p.x+dis[i][0];
next.y=p.y+dis[i][1];
if(inmap(next.x,next.y)) continue;
else if(map[next.x][next.y]=='.')
{
// 记录之前该点之前的点
path[next.x][next.y].x=p.x;
path[next.x][next.y].y=p.y;
path[next.x][next.y].step=1;
next.step=p.step+1;
visit[next.x][next.y]=1;
cd.push(next);
}
else
{
// 记录之前该点之前的点
path[next.x][next.y].x=p.x;
path[next.x][next.y].y=p.y;
path[next.x][next.y].step=map[next.x][next.y]-'0'+1;
next.step=p.step+map[next.x][next.y]-'0'+1;
visit[next.x][next.y]=1;
cd.push(next);
}
}
}
}
int main()
{
int i,j;
while(scanf("%d%d",&m,&n)!=EOF)
{
isfind=0;
for(i=0;i