BZOJ1619[Usaco2008 Nov]Guarding the Farm 保卫牧场

标签：zoj guard arm continue gre close clu nat turn

Description

The farm has many hills upon which Farmer John would like to place guards to ensure the safety of his valuable milk-cows. He wonders how many guards he will need if he wishes to put one on top of each hill. He has a map supplied as a matrix of integers; the matrix has N (1 < N <= 700) rows and M (1 < M <= 700) columns. Each member of the matrix is an altitude H_ij (0 <= H_ij <= 10,000). Help him determine the number of hilltops on the map. A hilltop is one or more adjacent matrix elements of the same value surrounded exclusively by either the edge of the map or elements with a lower (smaller) altitude. Two different elements are adjacent if the magnitude of difference in their X coordinates is no greater than 1 and the magnitude of differences in their Y coordinates is also no greater than 1.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: Line i+1 describes row i of the matrix with M space-separated integers: H_ij

Output

* Line 1: A single integer that specifies the number of hilltops

Sample Input

8 7
4 3 2 2 1 0 1
3 3 3 2 1 0 1
2 2 2 2 1 0 0
2 1 1 1 1 0 0
1 1 0 0 0 1 0
0 0 0 1 1 1 0
0 1 2 2 1 1 0
0 1 1 1 2 1 0

Sample Output

HINT

三个山丘分别是：左上角的高度为4的方格，右上角的高度为1的方格，还有最后一行中高度为2的方格．

bfs，找周围一圈的点，如果相同则加入队列，如果有高于本点的则不成为山丘；成为山丘的点要打上标记。

 1 #include<cstdio>
 2 using namespace std;
 3 int n,m,ans;
 4 int list[500000][2];
 5 int map[705][705];
 6 bool bo[705][705];
 7 int dx[8]={0,1,0,-1,1,-1,1,-1};
 8 int dy[8]={1,0,-1,0,1,1,-1,-1};
 9 int read(){
10     int x=0,f=1;char ch=getchar();
11     for(;ch<‘0‘||ch>‘9‘;ch=getchar())if(ch==‘-‘)f=-1;
12     for(;ch>=‘0‘&&ch<=‘9‘;ch=getchar())x=(x<<1)+(x<<3)+ch-‘0‘;
13     return x*f;
14 }
15 bool bfs(int sx,int sy){
16     int t=1,h=1;
17     bool boo=true;
18     list[1][0]=sx,list[1][1]=sy;
19     while(h<=t){
20         int nx=list[h][0],ny=list[h][1];
21         for(int i=0;i<=7;i++){
22             int x=nx+dx[i],y=ny+dy[i];
23             if(x<1||x>n||y<1||y>m)continue;
24             if(map[x][y]>map[nx][ny]){boo=false;continue;}
25             if(map[x][y]==map[nx][ny]&&!bo[x][y]){
26                 bo[x][y]=true;
27                 list[++t][0]=x;
28                 list[t][1]=y;
29             }
30         }
31         h++;
32     }
33     for(int i=1;i<=t;i++)list[i][0]=list[i][1]=0;
34     return boo;
35 }
36 int main(){
37     n=read(),m=read();
38     for(int i=1;i<=n;i++)
39         for(int j=1;j<=m;j++)
40             map[i][j]=read();
41     for(int i=1;i<=n;i++)
42         for(int j=1;j<=m;j++)
43             if(!bo[i][j])
44                 if(bfs(i,j))ans++;
45     printf("%d\n",ans);
46     return 0;
47 }

View Code

BZOJ1619[Usaco2008 Nov]Guarding the Farm 保卫牧场

标签：zoj guard arm continue gre close clu nat turn

原文地址：http://www.cnblogs.com/lightandshadow/p/7608626.html