救援行动：最短路径算法-优快云博客

探讨了如何使用最短路径算法从监狱中救出角色Angel。通过定义地图上的障碍物、道路及守卫，利用优先队列实现A*搜索算法来找出到达目标位置所需的最少时间。

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Rescue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37605 Accepted Submission(s): 13027

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

又是编译错误。

别人AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include<queue>
using namespace std;
struct node{
    int x,y,t;
    friend bool operator<(node n1,node n2){
        return n2.t<n1.t;
    }
};
int n,m;
char map[210][210];
int d[210][210];
int ax,ay,rx,ry;
node ft;
int mx[]={1,0,-1,0};
int my[]={0,-1,0,1};

int judge(int x,int y){
    if(x<0||x>=n||y<0||y>=m){
        return 0;
    }
    if(map[x][y]=='#'){
        return 0;
    }
    if(d[x][y]){
        return 0;
    }
    return 1;
}

void bfs(){
    priority_queue<node> q;
    ft.x=rx;
    ft.y=ry;
    ft.t=0;
    q.push(ft);
    d[rx][ry]=1;
    while(!q.empty()){
        ft=q.top();
        q.pop();
        //printf("%d,%d,%d\n",ft.x,ft.y,ft.t);
        int x=ft.x;
        int y=ft.y;
        if(x==ax&&y==ay){
            printf("%d\n",ft.t);
            return;
        }
        for(int i=0;i<4;i++){
            x=ft.x+mx[i];
            y=ft.y+my[i];
            if(!judge(x,y)){
                continue;
            }
            node nt;
            if(map[x][y]=='.'||map[x][y]=='a'){
                nt.x=x;
                nt.y=y;
                nt.t=ft.t+1;
                d[x][y]=1;
                q.push(nt);
            }else if(map[x][y]=='x'){
                nt.x=x;
                nt.y=y;
                nt.t=ft.t+2;
                d[x][y]=1;
                q.push(nt);
            }
        }
    }
    printf("Poor ANGEL has to stay in the prison all his life.\n");
}


int main()
{
    while(~scanf("%d%d",&n,&m)){
        memset(d,0,sizeof(d));
        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;
                }
            }
        }
        bfs();
    }
    return 0;
}

我的错误代码：

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m,sx,sy,ex,ey;
char map[201][201];
int vis[201][201];
int next[4][2]={0,1,0,-1,1,0,-1,0};

struct node
{
	int x,y,step;
	friend bool operator < (node a1,node a2)
	{
		return a1.step>a2.step;
	}
	
}b,c;

void bfs()
{
	b.x=sx;b.y=sy;b.step=0;
	priority_queue<node>q;
	q.push(b);
	vis[sx][sy]=1;
	while(!q.empty())
	{
		b=q.top();q.pop();
		if(b.x==ex&&b.y==ey)
		{
			printf("%d\n",b.step);return;
		}
		for(int i=0;i<4;i++)
		{
			c=b;
			c.x+=next[i][0];c.y+=next[i][1];
			if(c.x<0||c.y<0||c.x>=n||c.y>=m||vis[c.x][c.y]||map[c.x][c.y]=='#') continue;
			
			
				if(map[c.x][c.y]=='x')
				{
					vis[c.x][c.y]=1;
					c.step=b.step+2;
					q.push(c);
				}
				else
				{
					vis[c.x][c.y]=1;
					c.step=b.step+1;
					q.push(c);
				}
			
		}
	}
	printf("Poor ANGEL has to stay in the prison all his life.\n");return;
}

int main()
{
	while(~scanf("%d%d",&n,&m))
	{
	memset(vis,0,sizeof(vis));
		for(int i=0;i<n;i++)
	scanf("%s",map[i]);
	for(int i=0;i<n;i++)
	for(int j=0;j<m;j++)
	{
		if(map[i][j]=='r')
		sx=i,sy=j;
		else if(map[i][j]=='a')
		ex=i,ey=j;
	}
	
	bfs();
	}
	
}