Rescue 优先队列+广搜
2016-07-08 15:56
381 查看
Rescue
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 12 Accepted Submission(s) : 5
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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
给你r为起点,a为终点,#为墙壁,.为路径,遇到x则step+1,作优先队列自定义,优先step短的,然后广搜,每一次访问下一个结点取最小的step返回。
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#define N 205
using namespace std;
int vist
;
char map
;
int dir[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};
int n,m,k,sum,flag,sx,sy,ex,ey;
struct node
{
int x,y;
int step;
friend bool operator<(node a1,node a2)
{
return a2.step<a1.step;
}
};
int check(int a,int b)
{
if(a<0||a>=n||b<0||b>=m||map[a][b]=='#'||!vist[a][b])
return 1;
return 0;
}
int bfs()
{
int i;
priority_queue<node>Q;
node st,ed;
st.x=sx;
st.y=sy;
st.step=0;
vist[st.x][st.y]=1;
Q.push(st);
while(!Q.empty())
{
st=Q.top();
Q.pop();
if(st.x==ex&&st.y==ey)
return st.step;
for(i=0; i<4; i++)
{
ed=st;
ed.x=st.x+dir[i][0];
ed.y=st.y+dir[i][1];
if(check(ed.x,ed.y))
continue;
ed.step++;
if(map[ed.x][ed.y]=='x')
ed.step++;
if(vist[ed.x][ed.y]>ed.step)
{
vist[ed.x][ed.y]=ed.step;//取最小时间
Q.push(ed);
}
}
}
return 0;
}
int main()
{
int i,j;
while(~scanf("%d%d",&n,&m))
{
for(i=0; i<n; i++)
{
scanf("%s",map[i]);
for(j=0; map[i][j]; j++)
{
if(map[i][j]=='r')
{
sx=i,sy=j;
}
else if(map[i][j]=='a')
{
ex=i,ey=j;
}
}
}
memset(vist,1,sizeof(vist));
int ans=0;
ans=bfs();
if(ans)
printf("%d\n",ans);
else
printf("Poor ANGEL has to stay in the prison all his life.\n");
}
}
相关文章推荐
- GO中url.Values作为参数传递的两种解析方法
- HDU 1242 Rescue【BFS+优先队列】
- UIView的clipsToBounds属性和CALayer的masksToBounds属性的比较
- UIView的clipsToBounds属性和CALayer的masksToBounds属性的比较
- 3D UI场景中如何把XY平面的尺寸映射为屏幕像素
- JS中变量名作为if条件的 true/flase
- 虚拟现实-VR-UE4-编辑自定义Character-上下左右移动-旋转
- POJ2299 Ultra-QuickSor[树状数组+离散化 / 归并排序]
- UISearchController使用
- MySQL中char(36)被认为是GUID导致的BUG及解决方案
- EasyUI ComboGrid 集成分页、按键示例
- cannot find -luuid
- UITextFiled长度限制
- 2016.07.08,英语,《Vocabulary Builder》Unit 24
- 数据库连接池优化配置(druid,dbcp,c3p0)
- AngularUI Router
- Can't load native library. CPU arch invalid for this build
- hdu 5057 Argestes and Sequence(离线处理树状数组)
- 自定义Collection View布局
- Buileder(生成器)—对象创建型模式