洛谷 P2919 [USACO08NOV]守护农场Guarding the Farm

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.

题目大意：当一个点的相邻的八个点的高度都小于这个点时，这个点是个山丘。求山丘个数。

题解：搜索...这个叫灌水...将点的高度从高到低排序，进行dfs

代码：

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstring>

using namespace std;

int n,m,cnt,ans,h[][],vis[][];

struct Node{

    int x,y,h;

    bool operator < (const Node & a) const{return h>a.h;}

}node[];

void dfs(int x,int y){

    vis[x][y]=;

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

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

      if(x+i<||x+i>n||y+j<||y+j>m)continue;

      else if(h[x+i][y+j]<=h[x][y]&&!vis[x+i][y+j])dfs(x+i,y+j);

}

int main(){

    scanf("%d%d",&n,&m);

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

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

         scanf("%d",&node[++cnt].h);

         h[i][j]=node[cnt].h;

         node[cnt].x=i;node[cnt].y=j;

     }

     sort(node+,node+cnt+);

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

     if(!vis[node[i].x][node[i].y])dfs(node[i].x,node[i].y),ans++;

    cout<<ans<<endl;

    return ;

}