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
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
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 ",ans); 44 return 0; 45 } 46 47 int Aptal=Presist(); 48 int main(int argc,char*argv[]){;}