HDU 1072:Nightmare(BFS)
Nightmare
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6968 Accepted Submission(s): 3344
Given the layout of the labyrinth and Ignatius‘ start position, please tell Ignatius whether he could get out of the labyrinth, if he could, output the minimum time that he has to use to find the exit of the labyrinth, else output -1.
Here are some rules:
1. We can assume the labyrinth is a 2 array.
2. Each minute, Ignatius could only get to one of the nearest area, and he should not walk out of the border, of course he could not walk on a wall, too.
3. If Ignatius get to the exit when the exploding time turns to 0, he can‘t get out of the labyrinth.
4. If Ignatius get to the area which contains Bomb-Rest-Equipment when the exploding time turns to 0, he can‘t use the equipment to reset the bomb.
5. A Bomb-Reset-Equipment can be used as many times as you wish, if it is needed, Ignatius can get to any areas in the labyrinth as many times as you wish.
6. The time to reset the exploding time can be ignore, in other words, if Ignatius get to an area which contain Bomb-Rest-Equipment, and the exploding time is larger than 0, the exploding time would be reset to 6.
Each test case starts with two integers N and M(1<=N,Mm=8) which indicate the size of the labyrinth. Then N lines follow, each line contains M integers. The array indicates the layout of the labyrinth.
There are five integers which indicate the different type of area in the labyrinth:
0: The area is a wall, Ignatius should not walk on it.
1: The area contains nothing, Ignatius can walk on it.
2: Ignatius‘ start position, Ignatius starts his escape from this position.
3: The exit of the labyrinth, Ignatius‘ target position.
4: The area contains a Bomb-Reset-Equipment, Ignatius can delay the exploding time by walking to these areas.
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
4 -1 13
这道题判重方法比较特殊,因为一格可以反复的走,所以不能用bool判断,而是要根据炸弹剩余爆炸时间来判重,当走到一格的时候如果剩余时间小于等于之前在这个格子上的最大值,那么就不用把这个节点入队扩展状态了,因为这样做没有意义,除非剩余时间比原先的最大值要大,我们才会去将这个点加入队列然后更新改点剩余时间最大值。
#include<iostream> #include<cstring> #include<ctime> #include<algorithm> #include<cstdio> #include<queue> using namespace std; struct node { int x,y; int t; int res; }; int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; int sx, sy; int rest[10][10]; int g[10][10]; int n,m; bool jud(node &k) { if(k.x<0 || k.x>=n) return false; if(k.y<0 || k.y>=m) return false; if(g[k.x][k.y]==0) return false; if(k.res==0) return false; if(g[k.x][k.y]==4) k.res=6; if(k.res<=rest[k.x][k.y]) return false; rest[k.x][k.y]=k.res; return true; } int bfs() { queue<node> q; node a,k; int x,y; a.x=sx,a.y=sy,a.t=0,a.res=6; q.push(a); while(!q.empty()) { a=q.front(); q.pop(); for(int i=0; i<4; i++) { k.x = a.x + dir[i][0]; k.y = a.y + dir[i][1]; k.t = a.t + 1; k.res = a.res - 1; if(jud(k)) { if(g[k.x][k.y]==3) return k.t; q.push(k); } } } return -1; } int main() { int t; scanf("%d",&t); while(~scanf("%d%d",&n,&m)) { for(int i=0; i<n; i++) { for(int j=0; j<m; j++) { scanf("%d", &g[i][j]); if(g[i][j]==2) sx=i, sy=j; } } memset(rest,0,sizeof(rest)); rest[sx][sy]=6; printf("%d\n",bfs()); } return 0; }
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