HDU–1242 — Rescue [廣搜]

 

Rescue

 

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12409    Accepted Submission(s): 4541

 

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

 

 

Code:

 

一直WA, WA ,WA

原理是用了優先隊列而不是普通的隊列

這個原來用過

只是用了friend bool operator< ()

點這裡右面連結:  傳送門

 

 

#include<stdio.h>#include<string.h>#include<queue>using namespace std;typedef struct point{    int x,y;    int step;    friend bool operator< (point a,point b)//這裡這裡這裡    {        return a.step>b.step;//>號表示出隊的是優先順序低的,<號表示出隊優先順序高的    } }point;char a[211][211];int vis[211][211];int n,m,dir[4][2]={1,0,-1,0,0,1,0,-1};point angel;int r_x,r_y;int bfs(point p){    int i;priority_queue<point>q;    q.push(angel);    vis[angel.x][angel.y] = 1;    point head,temp    ;                 while(!q.empty())    {        head = q.top();    //出隊一個並對這個進行搜尋          q.pop();        if(head.x==r_x&&head.y==r_y) return head.step;//加上自己             for(i=0;i<4;i++)        {              temp.x = head.x + dir[i][0];//temp賦值              temp.y = head.y + dir[i][1];              if(temp.x<0 || temp.x>=n || temp.y<0 || temp.y>=m) continue;//越界                                  if(!vis[temp.x][temp.y] && a[temp.x][temp.y] != '#')                {                    if(a[temp.x][temp.y]=='x') temp.step = head.step + 2;//直接把到這一個地方的時間加上                    else temp.step = head.step + 1;                vis[temp.x][temp.y] = 1;                q.push(temp);        //判斷過後再入隊              }             }    }            return -1;}int main(){    int i,j,t;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=0;i<n;i++)        {            scanf("%s",a[i]);            for(j=0;j<m;j++)            {                if(a[i][j]=='a') { angel.x = i; angel.y = j; }                if(a[i][j]=='r') { r_x =i; r_y = j;}                           vis[i][j] = 0;            }        }            angel.step = 0;        t = bfs(angel);        if(t==-1) printf("Poor ANGEL has to stay in the prison all his life.\n");        else printf("%d\n",t);    }    return 0;}