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

* 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

* 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.

思路：搜索。

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int h[702][702];
int n,m,ans,H,tot; 
struct po{
    int x,y,h;
}a[703*703];
queue<int>qx,qy;
int fx,fy,rx,ry;
bool v[1000][1000];
int X[8]={0,0,1,-1,1,-1,1,-1};
int Y[8]={1,-1,0,0,1,1,-1,-1};
int cmp(po xx,po yy){
    return xx.h>yy.h;
}
void bfs(int x,int y){
    v[x][y]=1;
    qx.push(x);qy.push(y);
    while(!qx.empty()){
        fx=qx.front();qx.pop();
        fy=qy.front();qy.pop();
        H=h[fx][fy];
        for(int i=0;i<=7;i++){
            rx=X[i]+fx;ry=Y[i]+fy;
            if(rx<1||rx>n||ry<1||ry>m)    continue;
            if(v[rx][ry])    continue; 
            if(h[rx][ry]<=H){
                  qx.push(rx);qy.push(ry);
                  v[rx][ry]=1;    
            }
          }    
    }
}
int main(){ 
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){

            tot++;
            scanf("%d",&a[tot].h);
            h[i][j]=a[tot].h;
            a[tot].x=i;a[tot].y=j;
         } 
    sort(a+1,a+tot+1,cmp);
    memset(v,0,sizeof(v));
    for(int i=1;i<=tot;i++){
        int xx=a[i].x,yy=a[i].y;
        if(v[xx][yy])    continue;
        bfs(xx,yy);
        ans++; 
    }
    cout<<ans;
}