题意

Sol

贪心

从小到大枚举高度，把小于等于这一高度的相邻格子用并查集合并

那么这个集合内的所有格子都一定可以由这个集合内的一个最低点抽完水

那么合并之后（一定要在合并之后）

判断这一高度是否有城市，有则检查它所在的集合是否放了抽水机，没有就在这个集合中放一个

# include 
    
     # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))# define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)using namespace std;typedef long long ll;const int _(1005);IL int Input(){    RG int x = 0, z = 1; RG char c = getchar();    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}int n, m, ans, h[_][_], id[_][_], cnt, type[_][_], done[_ * _];int que[_ * _];int fa[_ * _], dx[4] = {1, -1}, dy[4] = {0, 0, 1, -1};struct Data{    int x, y, id;    IL bool operator <(RG Data B) const{        return h[x][y] < h[B.x][B.y];    }} p[_ * _];IL int Find(RG int x){    return fa[x] == x ? x : fa[x] = Find(fa[x]);}IL void Union(RG int Sx, RG int Sy, RG int fx){    for(RG int i = 0; i < 4; ++i){        RG int xx = Sx + dx[i], yy = Sy + dy[i];        if(xx < 1 || yy < 1 || xx > n || yy > m || h[xx][yy] > h[Sx][Sy]) continue;        RG int fy = Find(id[xx][yy]); fx = Find(fx);        if(fx == fy) continue;        done[fx] |= done[fy], fa[fy] = fx;    }}int main(RG int argc, RG char* argv[]){    n = Input(), m = Input();    for(RG int i = 1; i <= n; ++i)        for(RG int j = 1; j <= m; ++j){            h[i][j] = Input(), type[i][j] = h[i][j] > 0;            if(!type[i][j]) h[i][j] = -h[i][j];            id[i][j] = ++cnt, fa[cnt] = cnt, p[cnt] = (Data){i, j, cnt};        }    sort(p + 1, p + cnt + 1);    for(RG int i = 1, now; i <= cnt; ++i){        if(h[p[i].x][p[i].y] != now){            while(que[0]){                RG int x = Find(que[que[0]--]);                if(!done[x]) done[x] = 1, ++ans;            }            now = h[p[i].x][p[i].y];        }        Union(p[i].x, p[i].y, p[i].id);        if(type[p[i].x][p[i].y]) que[++que[0]] = p[i].id;    }    while(que[0]){        RG int x = Find(que[que[0]--]);        if(!done[x]) done[x] = 1, ++ans;    }    printf("%d\n", ans);    return 0;}