Description
Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of the bomb is set to 6 minutes. To prevent the bomb from exploding by shake, Ignatius had to move slowly, that is to move from one area to the nearest area(that is, if Ignatius stands on (x,y) now, he could only on (x+1,y), (x-1,y), (x,y+1), or (x,y-1) in the next minute) takes him 1 minute. Some area in the labyrinth contains a Bomb-Reset-Equipment. They could reset the exploding time to 6 minutes.Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.Output
For each test case, if Ignatius can get out of the labyrinth, you should output the minimum time he needs, else you should just output -1.Sample Input
3 3 3 2 1 1 1 1 0 1 1 3 4 8 2 1 1 0 1 1 1 0 1 0 4 1 1 0 4 1 1 0 0 0 0 0 0 1 1 1 1 4 1 1 1 3 5 8 1 2 1 1 1 1 1 4 1 0 0 0 1 0 0 1 1 4 1 0 1 1 0 1 1 0 0 0 0 3 0 1 1 1 4 1 1 1 1 1
Sample Output
4 -1 13
題目大意:
一個人在迷宮裡並且身上帶有定時炸彈,炸彈經過6秒就會爆炸,在這個迷宮裡面有5種屬性的方格,2代表出發點,3代表出口,1代表可以走的路,0代表牆,4代表在此點可以把當前的定時炸彈重置為6 . PS:在出口可以出去的條件是定時炸彈剩余時間大於0,同樣重置炸彈時間的條件是當前炸彈的爆炸時間大於0.
解題思路:
BFS+DP,當遇到下一步可以帶的時間比實際那一步多則更新爆炸時間。.
代碼:
#include#include #include #include using namespace std; int dirI[4]={1,0,-1,0},is,js,ie,je,n,m; int dirJ[4]={0,1,0,-1},timen[10][10]; struct node{ int i,j,t,v; node(int i0=0,int j0=0,int t0=0,int v0=0){ i=i0,j=j0,t=t0,v=v0; } }a[10][10]; void bfs(){ queue path; timen[is][js]=0; a[is][js].t=6; bool ans=false; path.push(a[is][js]); while(!path.empty()&&!ans){ node s=path.front(); path.pop(); for(int i=0;i<4;i++){ int di=s.i+dirI[i],dj=s.j+dirJ[i]; if(a[di][dj].v==0||di<0||dj<0||di>=n||dj>=m) continue; int temp=timen[s.i][s.j]+1,st=s.t-1 ; if(st>a[di][dj].t||timen[di][dj]==-1){ if(a[di][dj].v==4&&st>=1) {st=6;a[di][dj].t=st;timen[di][dj]=temp;path.push(a[di][dj]);} if(a[di][dj].v==1) {a[di][dj].t=st;timen[di][dj]=temp;path.push(a[di][dj]);} if(a[di][dj].v==3&&st>=1){ans=true;timen[di][dj]=temp;break;} } } } printf("%d\n",timen[ie][je]); } int main(){ int T; scanf("%d",&T); while(T--){ memset(timen,-1,sizeof(timen)); scanf("%d%d",&n,&m); for(int i=0;i