概述
Rescue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 27091 Accepted Submission(s): 9593
Problem Description
Angel 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.)
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.)
Input
First 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.
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.
Output
For 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
<span style="font-size:18px;">#include<cstdio> #include<cstring> #include<queue> using namespace std; struct node { int x,y; int temp; }; bool operator < (node a,node b) { return a.temp>b.temp;//按照从小到大输出 } char map[400][400]; int vis[400][400]; int n,m; int dx[4]={1,-1,0,0}; int dy[4]={0,0,1,-1}; int judge(int a,int b) { if(a<1||a>n||b<1||b>m||vis[a][b]||map[a][b]=='#') return 1; return 0; } void bfs(int sx,int sy) { priority_queue<node>que; node s; s.x=sx;s.y=sy; s.temp=0; que.push(s); while(!que.empty()) { node now; now=que.top(); //printf("%d %d %dn",now.x,now.y,now.temp); if(map[now.x][now.y]=='r') { printf("%dn",now.temp); return ; } que.pop(); for(int i=0;i<4;++i) { node end; end.x=now.x+dx[i]; end.y=now.y+dy[i]; if(judge(end.x,end.y)) continue; vis[end.x][end.y]=1; end.temp=now.temp+1; if(map[end.x][end.y]=='x') ++end.temp; que.push(end); } } printf("Poor ANGEL has to stay in the prison all his life.n"); return ; } int main() { int x,y;//记录出发的位置,我们的任务是天使找朋友 while(scanf("%d%d",&n,&m)!=EOF) { memset(vis,0,sizeof(vis)); int i,j; for(i=1;i<=n;++i) { scanf("%s",map[i]+1); for(j=1;j<=m;++j) { if(map[i][j]=='a') { x=i;y=j; } } } vis[x][y]=1; bfs(x,y); } return 0; } </span>
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