HDU 1242 Rescue (BFS+優先隊列)

標籤：hdu   bfs   

RescueTime Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 13   Accepted Submission(s) : 10Font: Times New Roman | Verdana |Georgia Font Size: ← →Problem DescriptionAngel 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.

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.

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.)
InputFirst line contains two integers stand for N and M.

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.

Process to the end of the file.
OutputFor 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."
Sample Input

7 8#.#####.#.a#..r.#..#x.....#..#.##...##...#..............

Sample Output

13因為有守衛的存在，所以先入隊的時間未必小，我們要輸出時間小的，就要先處理時間小的。所以用到了優先隊列。時間小的優先處理。#include <iostream>#include <stdio.h>#include <queue>#include <string.h>using namespace std;#define M 210struct node{int x,y,time; friend bool operator<(node a,node b)  //優先隊列自訂（我認為是這樣） {     return a.time>b.time;             //和sort相反a.time>b.time的時候b先入隊，也就是時間小的先入隊。 }};int n,m,vis[M][M];char map[M][M];int dis[4][2]={1,0,0,1,-1,0,0,-1};int DFS(int x,int y,int time){    int i,j;  node t,temp;  memset(vis,0,sizeof(vis));  t.x=x;t.y=y;t.time=time;  priority_queue<node>Q;                  //定義一個優先隊列。  Q.push(t);  vis[t.x][t.y]=1;  while(!Q.empty())  {      t=Q.top();      Q.pop();      if(map[t.x][t.y]==‘a‘)      {          return t.time;      }      for(i=0;i<4;i++)      {          temp.x=t.x+dis[i][0];          temp.y=t.y+dis[i][1];          temp.time=t.time+1;          if(temp.x>=0 && temp.x<n && temp.y>=0 && temp.y<m)           //先判斷越沒越界。            if(!vis[temp.x][temp.y] && map[temp.x][temp.y]!=‘#‘)       //再判斷合不合法，要不然會報錯。          {              if(map[temp.x][temp.y]==‘x‘)                             //越過守衛時間加一。              temp.time++;              vis[temp.x][temp.y]=1;              Q.push(temp);          }      }  }  return -1;}int main(){ int i,j,k; int fx,fy;  while(cin>>n>>m)  {      for(i=0;i<n;i++)      for(j=0;j<m;j++)      {        cin>>map[i][j];        if(map[i][j]==‘r‘)            fx=i,fy=j;      }      i=DFS(fx,fy,0);      if(i>=0)      printf("%d\n",i);      else        printf("Poor ANGEL has to stay in the prison all his life.\n");      } return 0;} 

HDU 1242 Rescue (BFS+優先隊列)