HDU 1242 - Rescue (BFS)

【搜索入门专题1】 hdu1242 J - Rescue c++ stl容器之优先队列+BFS Rescue 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 Angel's friends want to save Angel. Their task is: a 阅读详情

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32916    Accepted Submission(s): 11522


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.)
 

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
 

Author
CHEN, Xue
 

Source
 

POINT:
多起点的BFS。 遇X时间再多加1

#include <string>
#include <string.h>
#include <iostream>
#include <queue>
#include <math.h>
#include <stdio.h>
using namespace std;
#define LL long long
const int maxn = 222+55;
char mp[maxn][maxn];
int flag[maxn][maxn];
int n,m;
int ai,aj;
int dir[4][2]={0,1,0,-1,1,0,-1,0};
struct node
{
	int x,y,cnt;
	node(int xx,int yy,int cc):
	x(xx),y(yy),cnt(cc){}
};
int ans;
void bfs(int xxx,int yyy)
{
	queue<node> q;
	q.push(node(xxx,yyy,0));
	while(!q.empty()){
		int x=q.front().x;
		int y=q.front().y;
		int cnt=q.front().cnt;
		if(mp[x][y]=='a'){
			ans=min(cnt,ans);
			return;
		}
		q.pop();
		for(int i=0;i<4;i++){
			int xx=x+dir[i][0];
			int yy=y+dir[i][1];
			if(xx<1||xx>n||yy<1||yy>m||mp[xx][yy]=='#') continue;
			if(mp[xx][yy]=='x'){
				if(flag[xx][yy]!=0&&flag[xx][yy]<=cnt+2) continue;
				q.push(node(xx,yy,cnt+2));
				flag[xx][yy]=cnt+2;
			}else{
				if(flag[xx][yy]!=0&&flag[xx][yy]<=cnt+1) continue;
				q.push(node(xx,yy,cnt+1));
				flag[xx][yy]=cnt+1;
			}
		}
	}
}
int main()
{
	while(~scanf("%d %d",&n,&m)){
		ans=0x3f3f3f3f;
		for(int i=1;i<=n;i++){
			scanf("%s",mp[i]+1);
		}
		for(int i=1;i<=n;i++){
			for(int j=1;j<=m;j++){
				flag[i][j]=0;
				if(mp[i][j]=='a'){
					ai=i;
					aj=j;
				}
			}
		}
		for(int i=1;i<=n;i++){
			for(int j=1;j<=m;j++){
				if(mp[i][j]=='r'){
					bfs(i,j);
				}
			}
		}
		if(ans==0x3f3f3f3f) printf("Poor ANGEL has to stay in the prison all his life.\n");
		else printf("%d\n",ans);
	}
	
	return 0;
}


Simulink开发项1000例实战专栏--实例52:基于Simulink开发一个超导磁储能控制系统,实现对SMES系统的精确控制和优化 需求分析超导磁储能系统(Superconducting Magnetic Energy Storage, SMES)是一种利用超导线圈储存和释放电能的高效储能技术。SMES系统具有响应速度快、能量密度高、循环寿命长等优点,广泛应用于电力系统的调峰填谷、频率调节、电压稳定以及紧急备用电源等领域。本项目将基于Simulink平台,开发一个完整的超导磁储能控制系统,实现对SMES系统的精确控制和优化。我们将详细讲解如何设计和实现该系统,并结合实际应用案例进行说明。 阅读详情

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此题要注意的是多个。。r..还加优先队列。。 才能保证每次到r才是最小时间的。 Rescue Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5900Accepted Submission(s): 2193 Problem ...

weixin_30273931的博客 136

【BFS】hdu 1242 Rescue

优先队列+bfs求最短路

Gzbgreat的博客 384

bfs解决hdu1242Rescue

Rescue 题目链接 题目描述 具体请点击上方题目链接。题目大概意思就是给定一个n和m,然后给定一个由"#", “.” , “a” , “x” , “r” 构成的矩阵。然后要求从r到a的所需的最短的时间。 解题思路 需要注意的是"#“表示墙壁,不可通过,”."表示可通过,"a"表示要到达的目的地,"x"表示警官的位置,必须先耗费一定时间将将官杀掉,然后才能通过,“r"表示朋友所在位置,即开始的位置,但是题目中说"r” stands for each of Angel’s friend,说明可能会有好几个朋

weixin_43966538的博客 375

hdu1242 Rescue(BFS+优先队列)

原题链接 大意: 天使被MOLIGPY抓住了!他被Moligpy监禁了。监狱被描述为N * M(N,M &lt;= 200)矩阵。监狱里有WALL,ROAD和GUARD。 天使的朋友想要拯救天使。他们的任务是:接近天使。我们假设“接近天使”是为了达到安吉尔留下的位置。当网格中有一名后卫时,我们必须杀死他(或她?)才能进入网格。我们假设我们向上,向下,向右,向左移动需要1个单位时间,并且杀死一...

一叶之修的博客 294

HDU1242,Rescue(BFS)

一开始还没学BFS,前几次都是用DFS,但很明显,这样做会TLE,于是后来学了BFS之后,改用BFS,就AC了。 代码如下: #include<cstdio> #include<algorithm> #include<iostream> #include<cstring> #include<queue> using namespace s...

shamansi99的博客 271
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