杭电 HDU 1242 Rescue

http://acm.hdu.edu.cn/showproblem.php?pid=1242

问题:牢房里有墙(#),警卫(x)和道路( . ),天使被关在牢房里位置为a,你的位置在r处,杀死一个警卫要一秒钟,每走一步要一秒钟,求最短时间救出天使,不能救出则输出:Poor ANGEL has to stay in the prison all his life.  求最短路径,果断广搜BFS

限制及剪枝:
1、墙不能走,不能离开牢房范围
2、杀死一个警卫要多花一秒钟
3、当前步骤大于等于最短时间时不用继续再走(剪枝)
4、每次到达天使处要更新最短时间
5、不能走重复路(算剪枝把)          //这个也要说???囧。。。

AC代码:
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>

#define MAX 999999

using namespace std;

struct Node
{
    int x,y;
    int time;
};

char map[210][210];
int st[210][210];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int n,m,mintime;

void Bfs(Node p)
{
    Node now,next;
    queue<Node> q;
    int i;
    q.push(p);
    while(!q.empty())
    {
        now = q.front();
        q.pop();
        if(now.time >= mintime)       //剪枝,超过最短时间的不用再走了
        {
            break;
        }
        if(map[now.x][now.y] == 'a')  //找到天使时更新最短步骤
        {
            if(now.time < mintime)
            {
                mintime = now.time;
            }
        }
        if(map[now.x][now.y] == 'x')  //如果当前为门卫则多花1秒来杀死他
        {
            now.time += 1;
        }
        st[now.x][now.y] = 2;         //标记此路径已经到达
        next.time = now.time + 1;     //下一步的时间加一
        for(i = 0; i < 4; i++)
        {
            next.x = now.x + dir[i][0];
            next.y = now.y + dir[i][1];
            if(map[next.x][next.y] != '#' && st[next.x][next.y] == 0)
            {
                st[next.x][next.y] = 1;   //标记此路径已经入队
                q.push(next);
            }
        }
    }
    return ;
}

int main()
{
    int i,j;
    Node now;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(map,'#',sizeof(map));  //初始化全为墙
        memset(st,0,sizeof(st));      //初始化所有路径均未走过
        mintime = MAX;
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j <= m; j++)
            {
                cin >> map[i][j];
                if(map[i][j] == 'r')
                {
                    now.x = i;
                    now.y = j;
                    now.time = 0;
                }
            }
        }
        Bfs(now);
        if(mintime == MAX)
        {
            printf("Poor ANGEL has to stay in the prison all his life.
");
        }
        else
        {
            printf("%d
",mintime);
        }
    }

    return 0;
}


原文地址:https://www.cnblogs.com/javawebsoa/p/3202853.html