hdu 1026(优先队列+路径输出)
Ignatius and the Princess I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17171 Accepted Submission(s): 5495
Special Judge
Princess has been abducted by the BEelzebub feng5166, our hero Ignatius
has to rescue our pretty Princess. Now he gets into feng5166's castle.
The castle is a large labyrinth. To make the problem simply, we assume
the labyrinth is a N*M two-dimensional array which left-top corner is
(0,0) and right-bottom corner is (N-1,M-1). Ignatius enters at (0,0),
and the door to feng5166's room is at (N-1,M-1), that is our target.
There are some monsters in the castle, if Ignatius meet them, he has to
kill them. Here is some rules:
1.Ignatius can only move in four
directions(up, down, left, right), one step per second. A step is
defined as follow: if current position is (x,y), after a step, Ignatius
can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.
Your
task is to give out the path which costs minimum seconds for Ignatius
to reach target position. You may assume that the start position and the
target position will never be a trap, and there will never be a monster
at the start position.
input contains several test cases. Each test case starts with a line
contains two numbers N and M(2<=N<=100,2<=M<=100) which
indicate the size of the labyrinth. Then a N*M two-dimensional array
follows, which describe the whole labyrinth. The input is terminated by
the end of file. More details in the Sample Input.
each test case, you should output "God please help our poor hero." if
Ignatius can't reach the target position, or you should output "It takes
n seconds to reach the target position, let me show you the way."(n is
the minimum seconds), and tell our hero the whole path. Output a line
contains "FINISH" after each test case. If there are more than one path,
any one is OK in this problem. More details in the Sample Output.
.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.
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
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <stdlib.h>
using namespace std;
const int N = ;
struct Node{
int x,y,step,v;
};
bool operator < (Node a,Node b){
return a.step > b.step;
}
struct Per{
int x,y,v;
}cnt[N][N];
int n,m;
bool vis[N][N];
char graph[N][N];
int dir[][] = {{-,},{,},{,-},{,}};
int bfs(){
memset(vis,false,sizeof(vis));
Node s;
vis[][] = true;
s.x = ,s.y = ,s.step = ,s.v = ;
if(graph[][]=='X') return -;
if(graph[][]!='.'){
s.step = s.v = graph[][]-'';
}
priority_queue<Node> q;
q.push(s);
while(!q.empty()){
Node now = q.top();
q.pop();
if(now.x==n-&&now.y==m-) {
cnt[n][m].x = n-;
cnt[n][m].y = m-;
cnt[n][m].v = now.v;
return now.step;
}
for(int i=;i<;i++){
Node next;
next.x = now.x+dir[i][];
next.y = now.y+dir[i][];
if(vis[next.x][next.y]||next.x<||next.x>=n||next.y<||next.y>=m||graph[next.x][next.y]=='X') continue;
char temp = graph[next.x][next.y];
if(temp=='.'){
next.step = now.step+;
next.v = ;
}else{
next.step=now.step+temp-''+;
next.v = temp-'';
}
vis[next.x][next.y] = true;
q.push(next);
cnt[next.x][next.y].x = now.x;
cnt[next.x][next.y].y = now.y;
cnt[next.x][next.y].v = now.v;
}
}
return -;
}
int _count;
void dfs(int x,int y){
if(x==&y==) return;
dfs(cnt[x][y].x,cnt[x][y].y);
int v = cnt[x][y].v;
while(v--){
printf("%ds:FIGHT AT (%d,%d)\n",_count++,cnt[x][y].x,cnt[x][y].y);
}
if(x!=n&&y!=m)
printf("%ds:(%d,%d)->(%d,%d)\n",_count++,cnt[x][y].x,cnt[x][y].y,x,y);
}
int main(){
while(scanf("%d%d",&n,&m)!=EOF){
for(int i=;i<n;i++){
scanf("%s",graph[i]);
}
int ans = bfs();
if(ans==-){
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);
_count = ;
dfs(n,m);
}
printf("FINISH\n");
}
}
hdu 1026(优先队列+路径输出)的更多相关文章
- hdu 1026(BFS+输出路径) 我要和怪兽决斗
http://acm.hdu.edu.cn/showproblem.php?pid=1026 模拟一个人走迷宫,起点在(0,0)位置,遇到怪兽要和他决斗,决斗时间为那个格子的数字,就是走一个格子花费时 ...
- hdu 1026(Ignatius and the Princess I)BFS
Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (J ...
- [POJ] 1606 Jugs(BFS+路径输出)
题目地址:http://poj.org/problem?id=1606 广度优先搜索的经典问题,倒水问题.算法不需要多说,直接BFS,路径输出采用递归.最后注意是Special Judge #incl ...
- java实现将指定文件夹里所有文件路径输出到指定文件作为参数化文件给lr脚本使用
java实现将指定文件夹里所有文件路径输出到指定文件作为参数化文件给lr脚本使用 import java.io.BufferedReader; import java.io.BufferedWrite ...
- Floyd最短路(带路径输出)
摘要(以下内容来自百度) Floyd算法又称为插点法,是一种利用动态规划的思想寻找给定的加权图中多源点之间最短路径的算法,与Dijkstra算法类似. 该算法名称以创始人之一.1978年图灵奖获得者. ...
- 【CH5104】I-country 线性dp+路径输出
pre:在网格中,凸多边形可以按行(row)分解成若干段连续的区间 [ l , r ] ,且左端点纵坐标的值(col)满足先减后增,右端点纵坐标先增后减. 阶段:根据这个小发现,可以将阶段设置成每一行 ...
- URAL 1004 Sightseeing Trip(floyd求最小环+路径输出)
https://vjudge.net/problem/URAL-1004 题意:求路径最小的环(至少三个点),并且输出路径. 思路: 一开始INF开大了...无限wa,原来相加时会爆int... 路径 ...
- 洛谷 P2764 最小路径覆盖问题【最大流+拆点+路径输出】
题目链接:https://www.luogu.org/problemnew/show/P2764 题目描述 «问题描述: 给定有向图G=(V,E).设P 是G 的一个简单路(顶点不相交)的集合.如果V ...
- UVA--624 CD(01背包+路径输出)
题目http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...
随机推荐
- 剑桥offer(31~40)
31.题目描述 统计一个数字在排序数组中出现的次数. 思路:找到最低和最高,相减 class Solution { public: int GetNumberOfK(vector<int> ...
- run (牛客多校第二场)计数DP
链接:https://www.nowcoder.com/acm/contest/140/A来源:牛客网 题目描述 White Cloud is exercising in the playground ...
- mysql的concat用法
问题提出:mybatis的mapper文件中的模糊查询: mysql CONCAT()函数用于将多个字符串连接成一个字符串,是最重要的mysql函数之一,下面就将为您详细介绍mysql CONCAT( ...
- HTMLajax跨域向服务器写入数据
1.XMLHttpRequest升级版已经实现了跨域请求.不过需要在后台设置:header("Access-Control-Allow-Origin:http://www.a.com&quo ...
- [dhcpd]清除dhcp缓存
修改了dhcp的default-lease-time && max-lease-time 清除缓存: rm /var/lib/dhcpd.leases~ echo "&quo ...
- HDU1573 线性同余方程(解的个数)
X问题 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submiss ...
- python读书笔记-django架站过程总结(from the django book)
django架站过程总结:1.django-admin startproject store2.store这个project的目录下有:__init__,manage,setting,urls3.se ...
- JSP动态合并单元格
<%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core"%> <table ...
- mysql varchar到底能存多少字符。
utf8编码的varchar Mysql记录行数据是有限的.大小为64k,即65535个字节,而varchar要用1-2字节来存储字段长度,小于255的1字节,大于255的2字节. Mysql 5.0 ...
- 【洛谷 T47488】 D:希望 (点分治)
题目链接 看到这种找树链的题目肯定是想到点分治的. 我码了一下午,\(debug\)一晚上,终于做到只有两个点TLE了. 我的是不完美做法 加上特判\(A\)了这题qwq 记录每个字母在母串中出现的所 ...