HDU Search Practice Rescue

RescueTime limit:2000/1000ms (java/other) Memory limit:65536/32768k (Java/other) total submission (s): Accepted SUBM Ission (s): 13Problem Descriptionangel was caught by the moligpy! He was put into prison by Moligpy. The prison is described as a n * M (n, M &lt;=) matrix. There is WALLs, ROADs, and guards in the Prison.<br><br>angel ' s friends want to save Angel. Their task Is:approach Angel. We assume that &quot;approach angel&quot; IS-to-get to the position where Angel stays. When there's a guard in the grid, we must kill him (or his?) 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 has to calculate the minimal time to approach Ang El. (We can move only up, down, left and right, to the neighbor grid within bound, of course.) <br>

Inputfirst line contains-integers stand for N and M.<br><br>then N lines follows, every line have M charact ERs. "." stands for road, "a" stands for Angel, and "R" stands for each of the Angel ' s friend. <br><br>process to the end of the file.<br>

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 have to stay in the prison all his life." & Lt;br>

Sample Input7 8#.#####.#.a#. r.#. #x ..... #.##...##...#.............. Sample Output13 Simple test Instructions: The solution R arrives whether can reach a place, can, at least a few steps of thinking analysis: As with other topics, there is a target BFS

# include <iostream># include<cstring># include<queue># include<fstream>using namespacestd;structinfo{intx, y;} A, R;intN, M;Charmap[ -][ -];intdir[4][2] = {{-1,0},{1,0},{0, -1},{0,1},};intjishu[ -][ -];BOOLBorderintXinty) {    if(x >=1&& x <= n && y >=1&& y <=m)return true; return false;}intBFS ();intMain () {//fstream cin ("Aaa.txt");     while(Cin >> N >>m) { for(inti =1; I <= N; i++)        {             for(intj =1; J <= M; J + +) {cin>>Map[i][j]; if(Map[i][j] = ='a') {a.x=i; A.Y=J; }                if(Map[i][j] = ='R') {r.x=i; R.y=J; }            }        }        inthaha =BFS (); if(haha = =88888888) cout<<"Poor ANGEL have to stay in the prison all he life."<<Endl; Elsecout<< haha <<Endl; }    return 0;}intBFs () { for(inti =0; I < -; i++)     for(intj =0; J < -; J + +) Jishu[i][j]=88888888; JISHU[R.X][R.Y]=0; Queue<Info>Q;    Q.push (R);    Info T, Tep;  while(!Q.empty ()) {Tep=Q.front ();        Q.pop (); if(tep.x = = a.x && Tep.y = =a.y) { Break; }         for(inti =0; I <4; i++) {T.x= tep.x + dir[i][0]; T.y= Tep.y + dir[i][1]; if(!border (T.x, T.Y))//Out of Bounds, past, stone wall            Continue; if(Map[t.x][t.y] = ='#')            Continue; intmin; if(Map[t.x][t.y] = ='x') min= Jishu[tep.x][tep.y] +2; Elsemin= Jishu[tep.x][tep.y] +1; if(Min >=Jishu[t.x][t.y])Continue; JISHU[T.X][T.Y]=min; //cout << Jishu[t.x][t.y] << Endl;Q.push (t); }    }    returnjishu[a.x][a.y];}

HDU Search Practice Rescue