HDU 1026 Ignatius and the Princess I（优先队列+打印路径）

题意：n*m的迷宫，从（0,0）到（n-1,m-1），遇到怪物停留怪物所在方格中的数字个单位时间，求最短时间并打印路径；

思路：用bfs先搜最短路，搜最短路时一定要用优先队列，不然结果不对；在通过保存上一步的方法保存路径，到达终点时，将路径查询出来，遇到怪物是位置不变；

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
int n,m,s1,s2,e1,e2;
char mm[505][505];
int vis[505][505];
int  dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
struct node
{
    int x,y,t;
    bool operator<(const node&xx)const
    {
        return t>xx.t;
    }
};
struct lujin
{
    int x1,y1,x2,y2;
}record[505];
node pre[505][505];
int quan[505][505],route[505][505];
int bfs()
{
    priority_queue<node>q;
    vis[s1][s2]=1;
    node cur,now;
    node s;
    s.x=s1;s.y=s2;s.t=0;
    q.push(s);
    while(!q.empty())
    {
        cur=q.top();
        q.pop();
        for(int i=0;i<4;i++)
        {
            now.x=cur.x+dir[i][0];
            now.y=cur.y+dir[i][1];
            now.t=cur.t+1;
            if(now.x<0||now.x>=n||now.y<0||now.y>=m||vis[now.x][now.y]||mm[now.x][now.y]==‘X‘) continue;
            if(mm[now.x][now.y]>=‘1‘&&mm[now.x][now.y]<=‘9‘)
            {
                now.t+=mm[now.x][now.y]-‘0‘;
            }
            if(now.x==(n-1)&&now.y==(m-1))//到达终点，倒退回去找路径
            {
                int num=0,temp1,temp2;
                e1=now.x;e2=now.y;
                pre[now.x][now.y]=cur;
                while(e1!=0||e2!=0)//或的关系！！不是与，小bug调试了很久。。。
                {

                    if(mm[e1][e2]>=‘1‘&&mm[e1][e2]<=‘9‘)
                    {
                       for(int j=0;j<mm[e1][e2]-‘0‘;j++)
                       {
                           record[now.t-num].x1=pre[e1][e2].x;
                           record[now.t-num].y1=pre[e1][e2].y;
                           record[now.t-num].x2=e1;
                           record[now.t-num].y2=e2;
                           num++;
                       }
                    }
                        record[now.t-num].x1=pre[e1][e2].x;
                        record[now.t-num].y1=pre[e1][e2].y;
                        record[now.t-num].x2=e1;
                        record[now.t-num].y2=e2;
                        temp1=pre[e1][e2].x;
                        temp2=pre[e1][e2].y;
                        //printf("(%d,%d)->(%d,%d)\n",e1,e2,temp1,temp2);
                        e1=temp1;e2=temp2;
                    num++;
                }
                return now.t;
            }
            vis[now.x][now.y]=1;
            q.push(now);
            pre[now.x][now.y]=cur;//保存上一步
        }
    }
    return -1;
}
int main()
{
    int i,j,k,ans;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(mm,0,sizeof(mm));
        memset(vis,0,sizeof(vis));
        memset(quan,0,sizeof(quan));
        memset(pre,0,sizeof(pre));
        memset(route,0,sizeof(route));
        memset(record,0,sizeof(record));
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                scanf(" %c",&mm[i][j]);
                s1=0,s2=0;
            }
        }
        ans=bfs();
        if(ans==-1) printf("God please help our poor hero.\n");
        else printf("It takes %d seconds to reach the target position, let me show you the way.\n",ans);
        if(ans!=-1)
        {
            for(i=1;i<=ans;i++)
            {
                printf("%ds:",i);
                if(record[i-1].x2==record[i].x2&&record[i-1].y2==record[i].y2)
                    printf("FIGHT AT (%d,%d)\n",record[i].x2,record[i].y2);
                else
                printf("(%d,%d)->(%d,%d)\n",record[i].x1,record[i].y1,record[i].x2,record[i].y2);
            }
        }
        printf("FINISH\n");
    }
}