# BZOJ 1619: [Usaco2008 Nov]Guarding the Farm 保卫牧场【BFS】

## 1619: [Usaco2008 Nov]Guarding the Farm 保卫牧场

Time Limit: 5 Sec Memory Limit: 64 MB

Description

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.

Input

• 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

Output

• Line 1: A single integer that specifies the number of hilltops

Sample Input

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

Sample Output

3

HINT

#include<cstdio>
#include<queue>
#include<algorithm>
#define MAXN 705
#define par pair<int,int>
using namespace std;
const int f[8][2]={{1,1},{1,0},{1,-1},{0,1},{0,-1},{-1,1},{-1,0},{-1,-1}};
int n,m,Ans,a[MAXN][MAXN];
bool vis[MAXN][MAXN];
struct xcw{
int h,x,y;
bool operator <(const xcw b)const{return h<b.h;}
}P[MAXN*MAXN];
queue<par> que;
bool check(int x,int y,int t){
if(x<1||x>n||y<1||y>m) return 0;
if(vis[x][y]) return 0;
return a[x][y]<=t;
}
void BFS(int x,int y){
vis[x][y]=1;que.push(make_pair(x,y));
while(!que.empty()){
par p=que.front();que.pop();
x=p.first,y=p.second;
for(int i=0;i<8;i++)
if(check(x+f[i][0],y+f[i][1],a[x][y])) vis[x+f[i][0]][y+f[i][1]]=1,que.push(make_pair(x+f[i][0],y+f[i][1]));
}
}
int main(){
#ifndef ONLINE_JUDGE
freopen("prob.in","r",stdin);
freopen("prob.out","w",stdout);
#endif
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++) scanf("%d",&a[i][j]),P[(i-1)*m+j]=(xcw){a[i][j],i,j};
sort(P+1,P+1+n*m);
for(int i=n*m;i;i--)
if(!vis[P[i].x][P[i].y]) Ans++,BFS(P[i].x,P[i].y);
printf("%d\n",Ans);
return 0;
}
posted @ 2018-05-15 18:45  XSamsara  阅读(113)  评论(0编辑  收藏  举报