Ignatius and the Princess I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5989 Accepted Submission(s): 1816
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.
//HDU 1026
//光搜 + 优先队列,需要记录路径
#include<stdio.h>
#include<queue>
#include<iostream>
#include<string.h>
using namespace std;
const int MAXN=110;
struct node
{
int x,y;
int time;
friend bool operator<(node a,node b)//time小的优先级高
{
return a.time>b.time;
}
};
priority_queue<node>que; //优先队列
struct cmap
{
int nx,ny;//记录前驱,用来记录路径
char c;
}map[MAXN][MAXN];//记录map
int n,m;
int fight[MAXN][MAXN],mark[MAXN][MAXN];
int bfs()//从目标点开始搜索,搜索到起点。记录搜索的前驱,就记录下路径了
{
int k;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
node now,next;
while(!que.empty()) que.pop();
now.x=n-1;now.y=m-1;
if(map[now.x][now.y].c>='1'&&map[now.x][now.y].c<='9')
{
now.time=map[n-1][m-1].c-'0';
fight[now.x][now.y]=map[now.x][now.y].c-'0';
}
else now.time=0;
que.push(now);
while(!que.empty())
{
now=que.top();
que.pop();
if(now.x==0&&now.y==0) return now.time;
for(k=0;k<4;k++)
{
next.x=now.x+dir[k][0];
next.y=now.y+dir[k][1];
if(next.x>=0 && next.x<n && next.y>=0 &&next.y<m &&
!mark[next.x][next.y] &&map[next.x][next.y].c !='X')
{
if(map[next.x][next.y].c>='1'&&map[next.x][next.y].c<='9')
{
next.time = now.time+map[next.x][next.y].c-'0'+1;
fight[next.x][next.y]=map[next.x][next.y].c-'0';
}
else next.time=now.time+1;
que.push(next);
map[next.x][next.y].nx=now.x;
map[next.x][next.y].ny=now.y;
mark[next.x][next.y]=1;
}
}
}
return -1;
}
int main()
{
int i,j,flag,x,y,tx,ty,sec;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i<n;i++)
for(j=0;j<m;j++)
{
scanf(" %c",&map[i][j].c);//前面一个空格很关键
mark[i][j]=fight[i][j]=0;
}
mark[n-1][m-1]=1;
flag=bfs();
if(flag!=-1)
{
printf("It takes %d seconds to reach the target position, let me show you the way.\n",flag);
sec=1,x=y=0;
while(sec!=flag+1)
{
printf("%ds:(%d,%d)->(%d,%d)\n",sec++,x,y,map[x][y].nx,map[x][y].ny);
for(i=0;i<fight[map[x][y].nx][map[x][y].ny];i++)
printf("%ds:FIGHT AT (%d,%d)\n",sec++,map[x][y].nx,map[x][y].ny);
tx=map[x][y].nx;
ty=map[x][y].ny;
x=tx;y=ty;
}
}
else
printf("God please help our poor hero.\n");
printf("FINISH\n");
}
return 0;
}