洛谷—— P2919 [USACO08NOV]守护农场Guarding the Farm

https://www.luogu.org/problem/show?pid=2919

题目描述

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.

农场里有许多山丘，在山丘上约翰要设置哨岗来保卫他的价值连城的奶牛.

约翰不知道有多少山丘，也就不知道要设置多少哨岗.他有一张地图，用整数矩阵的方式描 述了农场N(1 <= N<=700)行M(1 < M<=700)列块土地的海拔高度好 H_ij (0 <= H_ij <= 10,000).请帮他 计算山丘的数量.

一个山丘是指某一个方格，与之相邻的方格的海拔高度均严格小于它.当然，与它相邻的方 格可以是上下左右的那四个，也可以是对角线上相邻的四个.

输入输出格式

输入格式：

 

  • 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

 

输出格式：

 

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

 

输入输出样例

输入样例#1：
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
输出样例#1：
3

说明

There are three peaks: The one with height 4 on the left top, one of the points with height 2 at the bottom part, and one of the points with height 1 on the right top corner.

 

DFS

每次从最高点向四周扩展、

 1 #include <algorithm>
 2 #include <cstdio>
 3 
 4 #define max(a,b) (a>b?a:b)
 5 inline void read(int &x)
 6 {
 7     x=0; register char ch=getchar();
 8     for(; ch>'9'||ch<'0'; ) ch=getchar();
 9     for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0';
10 }
11 const int N(705);
12 int fx[8]={-1,0,1,-1,1,-1,0,1};
13 int fy[8]={-1,-1,-1,0,0,1,1,1};
14 int n,m,H,ans,cnt,h[N][N];
15 struct Pos {
16     int x,y,h;
17     bool operator < (const Pos a)const
18     {
19         return h>a.h;
20     }
21 }pos[N*N];
22 bool vis[N][N];
23 void DFS(int x,int y)
24 {
25     vis[x][y]=1;
26     for(int tx,ty,i=0; i<8; ++i)
27     {
28         tx=x+fx[i]; ty=y+fy[i];
29         if(tx<1||ty<1||tx>n||ty>m) continue;
30         if(!vis[tx][ty]&&h[x][y]>=h[tx][ty]) DFS(tx,ty);
31     }
32 }
33 
34 int Presist()
35 {
36     read(n),read(m);
37     for(int i=1; i<=n; ++i)
38       for(int j=1; j<=m; ++j)
39         read(pos[++cnt].h),pos[cnt].x=i,pos[cnt].y=j,h[i][j]=pos[cnt].h;
40     std::sort(pos+1,pos+cnt+1);
41     for(int i=1; i<=cnt; ++i)
42         if(!vis[pos[i].x][pos[i].y]) DFS(pos[i].x,pos[i].y),ans++;
43     printf("%d\n",ans);
44     return 0;
45 }
46 
47 int Aptal=Presist();
48 int main(int argc,char*argv[]){;}

 

posted @ 2017-09-25 14:27  Aptal丶  阅读(194)  评论(0)  编辑  收藏  举报