题目传送门

题意：题意坑爹。问符合条件的的山顶个数

分析：降序排序后从每个点出发，假设为山顶，如果四周的点的高度>h – d那么可以走，如果走到已经走过的点且染色信息(山顶高度)不匹配那么就不是山顶。重点在于就算知道不是山顶也要染色完。

#include <bits/stdc++.h>
using namespace std;

const int N = 5e2 + 5;
const int INF = 0x3f3f3f3f;

int h, w, d;
struct Point    {
    int x, y, z;
    Point() {}
    Point(int x, int y, int z) : x (x), y (y), z (z) {}
    bool operator < (const Point &r) const  {
        return z > r.z;
    }
};
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
vector<Point> ps;
int mat[N][N];
int peak[N][N];
bool vis[N][N];

bool judge(int x, int y, int az)    {
    if (x < 1 || x > h || y < 1 || y > w || mat[x][y] <= az - d)   return false;
    else    return true;
}

int BFS(Point &a)    {
    if (vis[a.x][a.y])  return 0;
    vis[a.x][a.y] = true;   peak[a.x][a.y] = a.z;
    int cnt = 1;
    bool flag = true;
    queue<Point> que;   que.push (a);
    while (!que.empty ())   {
        Point r = que.front (); que.pop ();
        for (int i=0; i<4; ++i) {
            int tx = r.x + dx[i];
            int ty = r.y + dy[i];
            if (!judge (tx, ty, a.z))    continue;
            if (vis[tx][ty])    {
                if (peak[tx][ty] != a.z)    {
                    flag = false;
                }
                continue;
            }
            vis[tx][ty] = true; peak[tx][ty] = a.z;
            que.push (Point (tx, ty, mat[tx][ty]));
            if (mat[tx][ty] == a.z) cnt++;
        }
    }
    if (!flag)  cnt = 0;
    return cnt;
}

int run(void)   {
    memset (vis, false, sizeof (vis));
    memset (peak, 0, sizeof (peak));
    int ret = 0;
    for (int i=0; i<ps.size (); ++i)    {
        ret += BFS (ps[i]);
    }
    return ret;
}

int main(void)  {
    int T;  scanf ("%d", &T);
    while (T--) {
        scanf ("%d%d%d", &h, &w, &d);
        ps.clear ();
        for (int i=1; i<=h; ++i)    {
            for (int j=1; j<=w; ++j)    {
                scanf ("%d", &mat[i][j]);
                ps.push_back (Point (i, j, mat[i][j]));
            }
        }
        sort (ps.begin (), ps.end ());
        printf ("%d\n", run ());
    }

    return 0;
}