题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1026

Ignatius and the Princess I

5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX.
5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX1
5 6
.XX...
..XX1.
2...X.
...XX.
XXXXX.

It takes 13 seconds to reach the target position, let me show you the way.
1s:(0,0)->(1,0)
2s:(1,0)->(1,1)
3s:(1,1)->(2,1)
4s:(2,1)->(2,2)
5s:(2,2)->(2,3)
6s:(2,3)->(1,3)
7s:(1,3)->(1,4)
8s:FIGHT AT (1,4)
9s:FIGHT AT (1,4)
10s:(1,4)->(1,5)
11s:(1,5)->(2,5)
12s:(2,5)->(3,5)
13s:(3,5)->(4,5)
FINISH
It takes 14 seconds to reach the target position, let me show you the way.
1s:(0,0)->(1,0)
2s:(1,0)->(1,1)
3s:(1,1)->(2,1)
4s:(2,1)->(2,2)
5s:(2,2)->(2,3)
6s:(2,3)->(1,3)
7s:(1,3)->(1,4)
8s:FIGHT AT (1,4)
9s:FIGHT AT (1,4)
10s:(1,4)->(1,5)
11s:(1,5)->(2,5)
12s:(2,5)->(3,5)
13s:(3,5)->(4,5)
14s:FIGHT AT (4,5)
FINISH
God please help our poor hero.
FINISH

网上说用stack输出路径

可是stack还不会用，参照了大神的思路，只要倒着搜索就可以不用栈了，用二维数组记录每一步的前驱再往回打印就可以输出路径了；

本题用到了优先队列，时间小的先出队，对优先队列有疑惑的同学请移步 ：STL优先队列详解

【源代码】

#include <stdio.h>
#include <cstring>
#include <queue>
using namespace std;
#define maxn 110
char map[110][110];
bool vis[maxn][maxn];
struct point
{
    int x,y;int times;
    friend bool operator <(point a,point b)
    {
        return a.times>b.times;  //重载小于号，使得时间小的先出队列；
    }
};
struct Pre{
    int px,py;
}pre[maxn][maxn];
int r,c;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
bool inborder(point a)
{
    if(a.x<0||a.y<0||a.x>r-1||a.y>c-1)
        return false;
    return true;
}
void BFS(int x, int y)
{
    pre[x][y].px=-1;
    vis[x][y]=1;
    point st;
    st.x=x,st.y=y;
    if(map[st.x][st.y]!='.')
        st.times=(map[st.x][st.y]-'0');
    else
        st.times=0;
    priority_queue<point>Q;
    while(!Q.empty())
        Q.pop();
    Q.push(st);
    while(!Q.empty())
    {
        point now=Q.top();
        Q.pop();
        if(now.x==0&&now.y==0)
        {
            int key=1;
            printf("It takes %d seconds to reach the target position, let me show you the way.\n",now.times);
            while(pre[now.x][now.y].px!=-1)//循环找前驱，直到终点
            {
                int tx=pre[now.x][now.y].px;
                int ty=pre[now.x][now.y].py;
                if(map[tx][ty]=='.')
                {
                    printf("%ds:(%d,%d)->(%d,%d)\n",key++,now.x,now.y,tx,ty);
                }
                else//打怪耗时
                {
                    printf("%ds:(%d,%d)->(%d,%d)\n",key++,now.x,now.y,tx,ty);
                    int tt=map[tx][ty]-'0';
                    while(tt--)
                        printf("%ds:FIGHT AT (%d,%d)\n",key++,tx,ty);
                }
                now.x=tx;//将当前前驱作为起始
                now.y=ty;
            }
            //printf("%ds:(%d,%d)->(%d,%d)\n",key++,now.x,now.y,r-1,c-1);
            return ;
        }
        else
        {
            point next;
            for(int i=0;i<4;i++)
            {
                next.x=now.x+dx[i];
                next.y=now.y+dy[i];
                next.times=now.times;
                if(map[next.x][next.y]=='X'||vis[next.x][next.y]||!inborder(next))
                    continue;
                else
                {
                    vis[next.x][next.y]=1;
                    if(map[next.x][next.y]=='.')
                    {
                        next.times+=1;
                    }
                    else
                    {
                        next.times+=(map[next.x][next.y]-'0'+1);//除了打怪耗时，还有走到该坐标的耗时故+1；
                    }
                    pre[next.x][next.y].px=now.x;//记录前驱
                    pre[next.x][next.y].py=now.y;
                    Q.push(next);
                }
            }
        }
    }
    printf("God please help our poor hero.\n");
}
void init()
{
    memset(vis,0,sizeof(vis));
    memset(map,'\0',sizeof(map));
    memset(pre,0,sizeof(pre));
}
int main()
{
    //int r,c;
    while(~scanf("%d%d",&r,&c))
    {
        init();
        for(int i=0;i<r;i++)
            scanf("%s",map[i]);
        BFS(r-1,c-1);//从右下角搜索
        puts("FINISH");
    }
    return 0;
}

hdu1026 Ignatius and the Princess I(BFS+优先队列)

原文：http://blog.csdn.net/chaiwenjun000/article/details/45457541