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.
搜索
#include <algorithm> #include <cstdio> #include <queue> #define N 705 using namespace std; struct node { int h,x,y; bool operator<(node a)const { return h>a.h; } }sf[N*N]; bool vis[N][N]; int n,m,ans,tot,G[N][N],fx[8]={1,-1,0,0,1,1,-1,-1},fy[8]={0,0,-1,1,1,-1,-1,1}; struct Node { int x,y; }; void bfs(int x,int y) { queue<Node>Q; Q.push((Node){x,y}); vis[x][y]=1; for(Node now;!Q.empty();) { now=Q.front(); Q.pop(); int H=G[now.x][now.y]; for(int i=0;i<8;++i) { int tx=now.x+fx[i],ty=now.y+fy[i]; if(tx<1||tx>n||ty<1||ty>m||vis[tx][ty]||G[tx][ty]>H) continue; Q.push((Node){tx,ty}); vis[tx][ty]=1; } } } int Main() { scanf("%d%d",&n,&m); for(int i=1;i<=n;++i) { for(int j=1;j<=m;++j) { scanf("%d",&sf[++tot].h); sf[tot].x=i; sf[tot].y=j; G[i][j]=sf[tot].h; } } sort(sf+1,sf+1+tot); for(int i=1;i<=tot;++i) { if(vis[sf[i].x][sf[i].y]) continue; bfs(sf[i].x,sf[i].y); ans++; } printf("%d ",ans); return 0; } int sb=Main(); int main(int argc,char *argv[]){;}