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：

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、

 #include <algorithm>

 #include <cstdio>

 #define max(a,b) (a>b?a:b)

 inline void read(int &x)

 {

     x=; register char ch=getchar();

     for(; ch>''||ch<''; ) ch=getchar();

     for(; ch>=''&&ch<=''; ch=getchar()) x=x*+ch-'';

 }

 const int N();

 int fx[]={-,,,-,,-,,};

 int fy[]={-,-,-,,,,,};

 int n,m,H,ans,cnt,h[N][N];

 struct Pos {

     int x,y,h;

     bool operator < (const Pos a)const

     {

         return h>a.h;

     }

 }pos[N*N];

 bool vis[N][N];

 void DFS(int x,int y)

 {

     vis[x][y]=;

     for(int tx,ty,i=; i<; ++i)

     {

         tx=x+fx[i]; ty=y+fy[i];

         if(tx<||ty<||tx>n||ty>m) continue;

         if(!vis[tx][ty]&&h[x][y]>=h[tx][ty]) DFS(tx,ty);

     }

 }

 int Presist()

 {

     read(n),read(m);

     for(int i=; i<=n; ++i)

       for(int j=; j<=m; ++j)

         read(pos[++cnt].h),pos[cnt].x=i,pos[cnt].y=j,h[i][j]=pos[cnt].h;

     std::sort(pos+,pos+cnt+);

     for(int i=; i<=cnt; ++i)

         if(!vis[pos[i].x][pos[i].y]) DFS(pos[i].x,pos[i].y),ans++;

     printf("%d\n",ans);

     return ;

 }

 int Aptal=Presist();

 int main(int argc,char*argv[]){;}