490
HDU1242-Rescue
Rescue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16313 Accepted Submission(s): 5925
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
ZOJ Monthly, October 2003
題意：
救人，你在r，要救的人在a，'.'是路可以走，'#'是牆，'x'是警衛，走路花費一步，遇見警衛花費兩步。
問最小多少步能救人，能的話輸出步數，不能的話輸出"Poor ANGEL has to stay in the prison all his life."
思路：
搜索題，果斷用BFS，要注意的是，深搜回溯太多，效率不高，而簡單廣搜，入隊後每次出隊時，根據地
圖情況的不同，出隊元素所記憶的時間並不是層次遞增的，需要全部搜索才能找到正確答案。可以
根據時間進行優先性選擇，每次都要出隊當前隊列中步數最少的出隊，而入隊處理時，正常處理即可，出隊順序由優先隊列自動排序（小記錄在前）。這樣，我們就可以從“a”進行基於優先隊列的範圍搜索了，並且在第一
次抵達有朋友的位置時得到正確結果
AC代碼：
#include<stdio.h>
#include<string.h>
#include<functional>
#include<queue>
using namespace std;
char a[210][210];
int flag[210][210];
int next[4][2]={
{-1,0},
{1,0},
{0,-1},
{0,1}
};
int n,m;
struct node
{
int x,y;
int step;
bool operator < (const node &a) const {
return step>a.step;//最小值優先
}
};
void BFS(int x,int y)
{
priority_queue<node>q;
node q1,q2;
int i;
flag[x][y]=1;
q1.x=x;q1.y=y;
q1.step=0;
q.push(q1);
while(!q.empty())
{
q1=q.top();
q.pop();
for(i=0;i<4;i++)
{
q2.x=q1.x+next[i][0];
q2.y=q1.y+next[i][1];
q2.step=q1.step;
if(a[q2.x][q2.y]!='#'&&q2.x>=0&&q2.y>=0&&q2.x<n&&q2.y<m)
{
if(a[q2.x][q2.y]=='.'&&flag[q2.x][q2.y]==0)
{
flag[q2.x][q2.y]=1;
q2.step+=1;
q.push(q2);
continue;
}
if(a[q2.x][q2.y]=='x'&&flag[q2.x][q2.y]==0)
{
flag[q2.x][q2.y]=1;
q2.step+=2;
q.push(q2);
continue;
}
if(a[q2.x][q2.y]=='a')
{
printf("%d\n",q2.step+1);
return;
}
}
}
}
printf("Poor ANGEL has to stay in the prison all his life.\n");
}
int main()
{
int i,j,x,y;
while(scanf("%d %d",&n,&m)!=EOF)
{
memset(a,0,sizeof(a));
for(i=0;i<n;i++)
{
getchar();
for(j=0;j<m;j++)
{
scanf("%c",&a[i][j]);
if(a[i][j]=='r')
{
x=i;y=j;
}
}
}
memset(flag,0,sizeof(flag));
BFS(x,y);
}
return 0;
}
最後更新：2017-04-03 05:39:52