HDU 1242 Rescue（广搜，优先队列） HDU 1242 Rescue

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概述

HDU 1242 Rescue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21626 Accepted Submission(s): 7713

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

广搜，遇到守卫，时间 +2 ，进入队列，因为需要在队列中重新排序，因此，要用到优先队列！

这一题与 hdu 上面的 1180 诡异的楼梯属于同一类型！

附上hdu 1180 诡异楼梯链接：http://blog.csdn.net/xia842655187/article/details/47361835

附上源代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
using namespace std;
int b[4][2] = {1,0,0,1,-1,0,0,-1}; // 下右左上
char Map[250][250];
int visit[250][250];
int n,m;
struct A
{
int x,y,step;
bool operator < (const A &a) const
{
return a.step < step;
}
};
int dfs(int x1,int y1,int x2,int y2)
{
int sx,sy,t;
priority_queue<A> Q;
A e;
e.x = x1;
e.y = y1;
e.step = 0;
Q.push(e);
visit[x1][y1] = 1;
while(!Q.empty())
{
e = Q.top();
if(e.x == x2 && e.y == y2)
break;
Q.pop();
for(int i = 0;i < 4;i++)
{
sx = e.x + b[i][0];
sy = e.y + b[i][1];
if(sx >= 0 && sx <= n && sy >= 0 & sy <= m && (Map[sx][sy] == '.' || Map[sx][sy] == 'x' || Map[sx][sy] == 'a') && !visit[sx][sy])
{
if(Map[sx][sy] == 'x')
{
t = e.step + 2;
}
else
t = e.step + 1;
A e1;
e1.x = sx;
e1.y = sy;
e1.step = t;
Q.push(e1);
visit[sx][sy] = 1;
}
}
}
if(Q.empty())
return -1;
else
{
while(!Q.empty())
Q.pop();
return e.step;
}
}
int main()
{
while(cin >> n >> m)
{
int x1,x2,y1,y2;
memset(Map,'',sizeof(Map));
memset(visit,0,sizeof(visit));
for(int i = 0;i < n;i++)
{
scanf("%s",Map[i]);
for(int j = 0;j < m;j++)
{
if(Map[i][j] == 'a')
{
x2 = i;
y2 = j;
}
if(Map[i][j] == 'r')
{
x1 = i;
y1= j;
}
}
}
int ans = dfs(x1,y1,x2,y2);
if(ans >= 0)
cout << ans << endl;
else
cout << "Poor ANGEL has to stay in the prison all his life." << endl;
}
return 0;
}