# uva1660_ Patrol Robot (state BFS)

Source: Internet
Author: User

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tt2767
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Exercises
Build a three-dimensional array d[n][n][n], the first two dimensions is the coordinates, the third dimension is the blood bar, it means that when the blood bar is HP, whether it passes this. Use the structure to record the number of steps, the state output can be judged.

``#include <iostream>#include <algorithm>#include <cstdio>#include <cstdlib>#include <cstring>#include <queue>using namespace STD;Const Long DoublePI =ACOs(0.0) *2.0;Const intN = -;Const intDx[] = {0,1,0,-1};Const intDy[] = {-1,0,1,0};intN,m,k;structp{intx,y,hp,cnt;//hp is a blood bar, each pass barrier-1, CNT is the current number of stepsPintx =0,inty =0,intHP =0,intCNT =0): x (x), Y (y), hp (ph), CNT (CNT) {}};intD[n][n][n];//The ph value of the third dimension at the same time as this coordinateintG[n][n];intBFS ();intMain () {intcase;scanf("%d", &case);memset(g,-1,sizeof(G));//Remember to reset the map     while(case--) {scanf("%d%d%d", &m,&n,&k);//k is the maximum value of the blood bar         for(inti =0; I <m; i++) {GetChar ();//absorb ' \ n '             for(intj =0; J < N; J + +)scanf("%d", &g[i][j]); }printf("%d\n", BFS ()); }return  0;}intBFS () {memset(d,0,sizeof(d)); P S (0,0K0), G (M-1, N-1K0); d[0][0][K] =1; queue<P>Q Q.push (s); while(!q.empty ()) {p p = q.front (); Q.pop ();if(p.x = = g.x &&p.y = = g.y)returnp.cnt;//Reach the end of return steps        if(P.HP >=0)//have blood to keep walking{P tmp; for(inti =0; i<4;                i++) {tmp.x = p.x + dx[i];                TMP.Y = p.y + dy[i]; TMP.HP = G[tmp.x][tmp.y]? P.HP-1K//Met 0 o'clock full blood recoverytmp.cnt = p.cnt +1;if(0<= tmp.x && tmp.x < M &&0<= tmp.y && tmp.y <= n && d[tmp.x][tmp.y][tmp.hp] = =0&& tmp.hp >=0) {D[TMP.X][TMP.Y][TMP.HP] =1;                Q.push (TMP); }            }        }    }if(Q.empty ())return-1;//did not reach the end point}``

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uva1660_ patrol Robot (state BFS)

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