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、

 #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[]){;}

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