本文主要是介绍[ACM] HDU 5024 Wang Xifeng's Little Plot (构造,枚举),希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
Wang Xifeng's Little Plot
In the novel, Wang Xifeng was in charge of Da Guan Yuan, where people of Jia family lived. It was mentioned in the newly recovered pages that Wang Xifeng used to arrange rooms for Jia Baoyu, Lin Daiyu, Xue Baochai and other teenagers. Because Jia Baoyu was the most important inheritor of Jia family, and Xue Baochai was beautiful and very capable , Wang Xifeng didn't want Jia Baoyu to marry Xue Baochai, in case that Xue Baochai might take her place. So, Wang Xifeng wanted Baoyu's room and Baochai's room to be located at two ends of a road, and this road should be as long as possible. But Baoyu was very bad at directions, and he demanded that there could be at most one turn along the road from his room to Baochai's room, and if there was a turn, that turn must be ninety degree. There is a map of Da Guan Yuan in the novel, and redists (In China English, one whose job is studying 《Dream of the Red Chamber》is call a "redist") are always arguing about the location of Baoyu's room and Baochai's room. Now you can solve this big problem and then become a great redist.
There are several test cases.
For each case, the first line is an integer N(0<N<=100) ,meaning the map is a N × N matrix.
Then the N × N matrix follows.
The input ends with N = 0.
3 #.# ##. ..# 3 ... ##. ..# 3 ... ### ..# 3 ... ##. ... 0
3 4 3 5
N * N的矩阵, 走的方向为8个方向,其中' . '表示可走,下面用点表示,' #' 表示不可走,找出一条最长的路径包含几个点,输出点的个数,其中路径的要求是 最多包含一个直角(拐一个弯)。
int dir[8][2]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}} 定义方向, 按照 上,左上,右,右下,下,左下,左,左上 的顺序定义,编号为0,1,2,3,4,5,6,7
那么构成直角的两个方向有7对,分别是 0 2, 1 3 , 2 4, 3 5 ,4 6, 5 7 ,6 0 ,7 1
那么枚举每个可走的点,求出该点向8个方向分别走的最远距离,然后按照上面的配对,找出一条最长的合法路径。
枚举完所有可走的点,也就得出答案了。
代码:
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;char mp[102][102];
int n;
int dir[8][2]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};
int ds[8];//一个点每个方向最长可以走多远
int all[8];//每个点为直角的拐点两边最长走多远int main()
{while(scanf("%d",&n)!=EOF){if(n==0)break;getchar();for(int i=0;i<n;i++){for(int j=0;j<n;j++)scanf("%c",&mp[i][j]);getchar();}int ans=-1;for(int i=0;i<n;i++)for(int j=0;j<n;j++){if(mp[i][j]=='.'){memset(all,0,sizeof(all));memset(ds,0,sizeof(ds));for(int k=0;k<8;k++){int x=i,y=j;while(x>=0&&x<n&&y>=0&&y<n&&mp[x][y]=='.'){ds[k]++;//每个方向上最远走多少x+=dir[k][0];y+=dir[k][1];}}for(int i=0;i<8;i++){all[i]=ds[i]+ds[(i+2)%8];//构成直角的两个方向ans=max(ans,all[i]);//cout<<"ans"<<ans<<endl;}}}printf("%d\n",ans-1);}return 0;
}
这篇关于[ACM] HDU 5024 Wang Xifeng's Little Plot (构造,枚举)的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!