练习二1012


Rescue
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 74   Accepted Submission(s) : 22
[align=left]Problem Description[/align]
Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.<br><br>Angel's friends want to save Angel. Their task
is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time,
and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.<br><br>You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)<br>
 

[align=left]Input[/align]
First line contains two integers stand for N and M.<br><br>Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend. <br><br>Process to the end of the
file.<br>
 

[align=left]Output[/align]
For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life." <br>
 

[align=left]Sample Input[/align]

7 8<br>#.#####.<br>#.a#..r.<br>#..#x...<br>..#..#.#<br>#...##..<br>.#......<br>........<br>

 

[align=left]Sample Output[/align]

13<br>

 

[align=left]Author[/align]
CHEN, Xue
 

[align=left]Source[/align]
ZOJ Monthly, October 2003
 

Statistic |

Submit |
题意：
在一张地图中，a被困住，r去救a，r不止一个，x代表敌人，r每走一步用时为1，杀掉敌人要格外加1，问最短时间。
思路：
是一个优先队列+bfs，以时间为标准优先，搜索时可以从a开始去搜索r,因为a只有一个，r有多个。没看答案，自己写了好多遍终于过了。注意下面写的注释。
代码：
#include<iostream>

#include<string.h>

#include<queue>

using namespace std;

int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}},n,m,sx,sy;

char map[201][201];

bool vis[201][201];

bool inmap(int x,int y)

{

 if(x>=1&&x<=n&&y>=1&&y<=m)

  return true;

 return false;

}

struct point

{

 int x,y,tim;

 friend bool operator<(point a,point b)

 {

  return a.tim>b.tim;

 }

}p1,p2;

int bfs(int a,int b,int c)

{

 p1.x=a;

 p1.y=b;

 p1.tim=c;

 priority_queue<point>q;

 q.push(p1);

 while(!q.empty())

 {

  p1=q.top();

  q.pop();

  if(map[p1.x][p1.y]=='r')

   return p1.tim;

  for(int i=0;i<4;i++)

  {

   int n_x=p1.x+dir[i][0];

      int n_y=p1.y+dir[i][1];

   if(!inmap(n_x,n_y)||vis[n_x][n_y]||map[n_x][n_y]=='#')

    continue;

   if(map[n_x][n_y]=='x')

   {

    vis[n_x][n_y]=1;

    p2.x=n_x;

    p2.y=n_y;

    p2.tim=p1.tim+2;

    q.push(p2);

   }

   if(map[n_x][n_y]=='r'||map[n_x][n_y]=='.'){

   vis[n_x][n_y]=1;

   p2.x=n_x;

   p2.y=n_y;

   p2.tim=p1.tim+1;

   q.push(p2);

   }

  }

 }

 return -1;

}

int main()

{

 while(cin>>n>>m)

 {

  for(int i=1;i<=n;i++){

   for(int j=1;j<=m;j++){

    cin>>map[i][j];

   }

  }

  for(int k=1;k<=n;k++){

   for(int j=1;j<=m;j++){

    if(map[k][j]=='a')

    {sx=k;sy=j;}

   }

  }

  memset(vis,0,sizeof(vis));

  vis[sx][sy]=1;

  int cnt=bfs(sx,sy,0);                                                                                          //有返回值的函数一定要注意第一次调用后再调用就不对了。（if(bfs(....)) cout<<bfs(....)...）就不

  if(cnt==-1) cout<<"Poor ANGEL has to stay in the prison all his life."<<endl; // 对。

  else cout<<cnt<<endl;

 }

 return 0;

}