Rescue
2016-07-29 18:19
337 查看
B - Rescue
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status
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.)
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.
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
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status
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.)
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.
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
#include<stdio.h> #include<string.h> #include<queue> #include<iostream> #include<algorithm> using namespace std; const int MAXN = 213; int vis[MAXN][MAXN]; char map[MAXN][MAXN]; int dis[4][2] = {0,1,1,0,0,-1,-1,0}; int m,n,ans,cnt,ax,ay,rx,ry; struct node{ int x,y,time; bool operator < (const node & a) const { return time > a.time; } }e,ee; priority_queue<node>q; int bfs(int x,int y){ node e,ee; e.x = x; e.y = y;e.time = 0; q.push(e); while(!q.empty() ){ e = q.top () ;q.pop() ; for(int i=0; i<4; i++){ int nx = e.x + dis[i][0]; int ny = e.y + dis[i][1]; if(nx>=0 && ny>=0 && nx<n && ny<m && !vis[nx][ny] && map[nx][ny] != '#'){ if(map[nx][ny] == '.' || map[nx][ny] == 'a'){ ee.x = nx ; ee.y = ny ; ee.time = e.time + 1; if (ee.x == ax && ee.y == ay) return ee.time; q.push(ee); vis[ee.x][ee.y] = 1; } if(map[nx][ny] == 'x'){ ee.x = nx ; ee.y = ny ; ee.time = e.time + 2; q.push(ee); vis[ee.x][ee.y] = 1; } } } } return 0; } int main(){ while(scanf("%d %d", &n, &m) != EOF){ ans = 0; ax = ay = rx = ry = 0; for(int i=0; i<n; i++){ scanf("%s",&map[i]); for(int j=0; j<m; j++){ if(map[i][j] == 'a'){ ax = i; ay = j; } if(map[i][j] == 'r'){ rx = i; ry = j; } } } memset(vis,0,sizeof(vis)); while(!q.empty() ) q.pop(); vis[rx][ry] = 1; ans = bfs(rx,ry); if(ans) cout<<ans<<endl; else printf("Poor ANGEL has to stay in the prison all his life.\n"); } return 0; }
相关文章推荐
- EasyUI 组合网格
- 给手势UITapGestureRecognizer绑定tag
- EasyUI 组合框
- java设计模式——建造者模式(Builder Pattern)
- EasyUI 组合树
- EasyUI 组合
- Android 5.1 SystemUI 状态栏修改
- EasyUI 文本框、文件框和开关按钮
- ueditor 文本编辑器
- 98.Which two statements are true regarding the usage of the SQL*Loader utility? (Choose two.)
- Rescue hd 1242
- POJ 2031 Building a Space Station (最小生成树)
- EasyUI 验证框
- EasyUI 表单
- 蓝牙bluetooth之二-源码分布
- Android UI(ToggleButton)详解
- EasyUI 提示框
- EasyUI 进度条
- EasyUI 搜索框
- UIAlertController