HDU 1206 Ignatius and the Princess I(BFS+记录路径)
来源:互联网 发布:网站建设优化公司 编辑:程序博客网 时间:2024/06/03 17:22
Ignatius and the Princess I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18847 Accepted Submission(s): 6090
Special Judge
Problem Description
The 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.
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
The 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.
Output
For 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.
Sample Input
5 6.XX.1...X.2.2...X....XX.XXXXX.5 6.XX.1...X.2.2...X....XX.XXXXX15 6.XX.....XX1.2...X....XX.XXXXX.
Sample Output
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)FINISHIt 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)FINISHGod please help our poor hero.FINISH
Author
Ignatius.L
Recommend
We have carefully selected several similar problems for you: 1072 1175 1180 1240 1043
记录一下
用优先队列和方向记录路径
#include <stdio.h> #include <string.h> #include <queue> using namespace std; struct node{ int x,y,step; friend bool operator<(node n1,node n2){ return n2.step<n1.step; } }; int map[105][105]; int flag[105][105]; int cnt[105][105];//记录怪的血量 int to[4][2]={1,0,-1,0,0,1,0,-1}; int n,m,tim; int check(int x,int y){ if(x<0||y<0||x>=n||y>=m) return 1; if(map[x][y]==-1) return 1; return 0; } int bfs(){ int i; priority_queue<node>Q; node a,next; a.x=0; a.y=0; a.step=0; map[0][0]=-1; Q.push(a); while(!Q.empty()){ a=Q.top(); Q.pop(); if(a.x==n-1&&a.y==m-1) return a.step; for(i=0;i<4;i++){ next=a; next.x+=to[i][0]; next.y+=to[i][1]; if(check(next.x,next.y)) continue; next.step=a.step+map[next.x][next.y]+1;//记录时间 map[next.x][next.y] = -1;//标记为走过 flag[next.x][next.y] = i+1;//记录朝向 Q.push(next); } } return 0; } void print(int x,int y){ int n_x,n_y; if(!flag[x][y]) return; n_x=x-to[flag[x][y]-1][0]; n_y=y-to[flag[x][y]-1][1]; print(n_x,n_y); printf("%ds:(%d,%d)->(%d,%d)\n",tim++,n_x,n_y,x,y); while(cnt[x][y]--){ printf("%ds:FIGHT AT (%d,%d)\n",tim++,x,y); } } int main(){ int i,j; while(~scanf("%d%d",&n,&m)){ memset(map,0,sizeof(map)); memset(flag,0,sizeof(flag)); memset(cnt,0,sizeof(cnt)); char s[105]; for(i=0;i<n;i++){ scanf("%s",s); for(j=0;s[j];j++){ if(s[j]=='.') map[i][j]=0; else if(s[j]=='X') map[i][j]=-1; else map[i][j]=cnt[i][j]=s[j]-'0'; } } int ans=0; ans=bfs(); if(ans){ printf("It takes %d seconds to reach the target position, let me show you the way.\n",ans); tim=1; print(n-1,m-1); } else printf("God please help our poor hero.\n"); printf("FINISH\n"); } return 0; }
0 0
- HDU 1026 Ignatius and the Princess I && BFS+记录路径
- HDU 1026 Ignatius and the Princess I(BFS+记录路径)
- HDU 1206 Ignatius and the Princess I(BFS+记录路径)
- hdu1026 Ignatius and the Princess I(BFS+路径记录)
- HDU 1026Ignatius and the Princess I(bfs+记录路径)
- HDU 1026 Ignatius and the Princess I (BFS+记录路径)
- HDU 1026 Ignatius and the Princess I(bfs +记录路径)
- HDU 1026 Ignatius and the Princess I(BFS+优先队列+路径记录)
- HDU 1026 Ignatius and the Princess I (bfs + 优先队列 + 路径记录)
- HDU1026 Ignatius and the Princess I 【BFS】+【路径记录】
- HDU 1026 Ignatius and the Princess I BFS打印路径
- hdu 1026 Ignatius and the Princess I bfs 优先队列 路径记录
- hdu1026 Ignatius and the Princess I(bfs+路径)
- 杭电1026 Ignatius and the Princess I(BFS+路径记录)
- 杭电-1026Ignatius and the Princess I(BFS+记录路径)
- HDOJ 1026 Ignatius and the Princess I (BFS+优先队列+记录路径)
- HDU-1026 Ignatius and the Princess I (BFS)
- hdu 1026 Ignatius and the Princess I(BFS)
- 二叉树的后序遍历
- LintCode 二叉树的所有路径
- Win7中开发第一个Storm程序
- 如何配置OpenVPN?
- Redis C 语言客户端 hiredis 的使用
- HDU 1206 Ignatius and the Princess I(BFS+记录路径)
- session入memcache
- Java自带性能分析工具
- go语言坑之并发访问map
- Flume Kafka收集Docker容器内分布式日志应用实践
- nginx实践(一)、安装和部署
- 【SSLGZ 2648】线段树练习五
- L2-023. 图着色问题
- 告诉您:解决Windows 10上的广告办法