洛谷 P2919 [USACO08NOV]守护农场Guarding the Farm

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.

搜索

屠龙宝刀点击就送

#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[]){;}