HDU 2653Waiting ten thousand years for Love
来源:互联网 发布:什么软件有哥特式字体 编辑:程序博客网 时间:2024/05/29 10:12
Problem Description
It was ten thousand years, after Demon Lemon caught Yifenfei’s love. In order to revenge and save his love, Yifenfei have been practising sword all day long and his Kongfu skills becomes so powerful that he can kill Demon Lemon immediately. Recently, Yifenfei have found Lemon’s castle, and now he is going to kill Lemon. At the same time, hearing about the terrible news, Demon Lemon is now preparing for escaping...
Now Yifenfei has got the map of the castle.
Here are all symbols of the map:
Only one ‘Y’ Indicates the start position of Yifenfei.
Only one ‘L’ Indicates the position of Demon Lemon.
‘.’ Indicate the road that Yifenfei can walk on it, or fly over it.
‘#’ Indicate the wall that Yifenfei can not walk or flay through it.
‘@’ Indicate the trap that Yifenfei can not walk on it, but can fly over it.
Yifenfei can walk or fly to one of up, down, left or right four directions each step, walk costs him 2 seconds per step, fly costs him 1 second per step and 1 magic power. His magic power will not increased, and if his magic power is zero, he can not fly any more.
Now Yifenfei asks you for helping him kill Demon Lemon smoothly. At the same time, Demon Lemon will Leave the castle Atfer T seconds. If Yifenfei can’t kill Demon Lemon this time, he have to wait another ten thousand years.
Now Yifenfei has got the map of the castle.
Here are all symbols of the map:
Only one ‘Y’ Indicates the start position of Yifenfei.
Only one ‘L’ Indicates the position of Demon Lemon.
‘.’ Indicate the road that Yifenfei can walk on it, or fly over it.
‘#’ Indicate the wall that Yifenfei can not walk or flay through it.
‘@’ Indicate the trap that Yifenfei can not walk on it, but can fly over it.
Yifenfei can walk or fly to one of up, down, left or right four directions each step, walk costs him 2 seconds per step, fly costs him 1 second per step and 1 magic power. His magic power will not increased, and if his magic power is zero, he can not fly any more.
Now Yifenfei asks you for helping him kill Demon Lemon smoothly. At the same time, Demon Lemon will Leave the castle Atfer T seconds. If Yifenfei can’t kill Demon Lemon this time, he have to wait another ten thousand years.
Input
Lots of test cases, please process to end of file. In each test case, firstly will have four integers N, M, T, P(1 <= N, M, P <= 80, 1 <= T <= 100000), indicates the size of map N * M, the Lemon’s leaving time(after T seconds, Lemon will disappeared) and Yifenfei’s magic power. Then an N * M two-dimensional array follows indicates the map.
Output
For each test case, first Print a line “Case C:”, C indicates the case number. Then, if Yifenfei can kill Demon Lemon successfully, Print “Yes, Yifenfei will kill Lemon at T sec.”, T indicates the minimum seconds he must cost. Otherwise, Print ”Poor Yifenfei, he has to wait another ten thousand years.”
Sample Input
2 3 2 2Y@L###2 3 4 1Y@L###2 3 4 0Y.L###2 3 3 0Y.L###
Sample Output
Case 1:Yes, Yifenfei will kill Lemon at 2 sec.Case 2:Poor Yifenfei, he has to wait another ten thousand years.Case 3:Yes, Yifenfei will kill Lemon at 4 sec.Case 4:Poor Yifenfei, he has to wait another ten thousand years.HintHintCase 1: Yifenfei cost 1 second and 1 magic-power fly to ‘@’, but he can not step on it, he must cost another 1 second and 1 magic-power fly to ‘L’ and kill Lemon immediately. Case 2: When Yifenfei Fly to ‘@’, he has no power to fly,and is killed by trap.
直接bfs即可,判断一下格子的状态。
#include<queue>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define rep(i,j,k) for (int i=j;i<=k;i++)const int INF=0x7FFFFFFF;const int sz=100;int f[sz][sz][sz],mp[sz][sz];int n,m,g,t,bx,by,ex,ey,T=0;char s[sz][sz];int x[4]={0,0,1,-1};int y[4]={1,-1,0,0};struct point{ int x,y,z; point(int x=0,int y=0,int z=0):x(x),y(y),z(z){};};int main(){ while (~scanf("%d%d%d%d",&n,&m,&t,&g)) { rep(i,1,n) rep(j,1,m) rep(k,0,g) f[i][j][k]=INF; memset(mp,0,sizeof(mp)); rep(i,1,n) { scanf("%s",s[i]+1); rep(j,1,m) { if (s[i][j]=='Y') {s[i][j]='.'; bx=i; by=j; } if (s[i][j]=='L') {s[i][j]='.'; ex=i; ey=j; } mp[i][j]=s[i][j]=='.'?1:s[i][j]=='@'?2:0; } } queue<point> p; p.push(point(bx,by,0)); f[bx][by][0]=0; while (!p.empty()) { point q=p.front(); p.pop(); if (q.z > g || f[q.x][q.y][q.z] > t || q.x==ex&&q.y==ey) continue; rep(i,0,3) { int X=q.x+x[i],Y=q.y+y[i]; if (!mp[X][Y]) continue; if (mp[q.x][q.y]==1&&mp[X][Y]==1) { if (f[X][Y][q.z]>f[q.x][q.y][q.z]+2) { f[X][Y][q.z]=f[q.x][q.y][q.z]+2; p.push(point(X,Y,q.z)); } if (f[X][Y][q.z+1]>f[q.x][q.y][q.z]+1) { f[X][Y][q.z+1]=f[q.x][q.y][q.z]+1; p.push(point(X,Y,q.z+1)); } } else if (f[X][Y][q.z+1]>f[q.x][q.y][q.z]+1) { f[X][Y][q.z+1]=f[q.x][q.y][q.z]+1; p.push(point(X,Y,q.z+1)); } } } int flag=INF; rep(i,0,g) flag=min(flag,f[ex][ey][i]); printf("Case %d:\n",++T); if (flag<=t) printf("Yes, Yifenfei will kill Lemon at %d sec.\n",flag); else printf("Poor Yifenfei, he has to wait another ten thousand years.\n"); } return 0;}
0 0
- HDU 2653 Waiting ten thousand years for Love
- HDU 2653 Waiting ten thousand years for Love
- HDU 2653Waiting ten thousand years for Love
- Waiting ten thousand years for Love
- Hdu 2579 Dating with girls(2) && hdu 2653 Waiting ten thousand years for Love【Bfs】
- hdoj 2653 Waiting ten thousand years for Love(优先队列+BFS)
- HDU2653 Waiting ten thousand years for Love (三维广搜+优先队列)
- HDU 2619 - Love you Ten thousand years (数论)
- hdu 2619 Love you Ten thousand years(数论,待解决)
- The Star Zoo——5、Waiting for fifteen thousand years
- ten beautiful sentences for love
- Concurrent Ten Thousand Connections.
- A thousand years
- Ten years. One vision
- 十年,ten years
- The Recent Ten Years
- lessen ten years
- 十年 Ten years
- window 上安装 Cygwin 和 SSH 服务
- 很纠结的问题:org.apache.axis2.AxisFault: 服务器未能识别 HTTP 头 SOAPAction 的值
- C++内存泄露和检测
- iOS图片轮播器(第三方SDCycleScrollView)
- dpkg依赖错误
- HDU 2653Waiting ten thousand years for Love
- 解决pod安装 requires Ruby version >= 2.2.2
- 内存泄漏问题
- 懒加载(LAZY)在View层报错的解决方法: OpenSessionInViewFilter
- linux Sybase ASE 查询当前版本和所有数据库
- C语言入门(二十)结构体
- Express Error: No default engine was specified and no extension was provided
- myeclipse加载maven工程提示pom.xml无法解析!
- LeetCode 231. Power of Two