F. Polycarp and Hay

題目連線：

Description

The farmer Polycarp has a warehouse with hay, which can be represented as an n × m rectangular table, where n is the number of rows, and m is the number of columns in the table. Each cell of the table contains a haystack. The height in meters of the hay located in the i-th row and the j-th column is equal to an integer ai, j and coincides with the number of cubic meters of hay in the haystack, because all cells have the size of the base 1 × 1. Polycarp has decided to tidy up in the warehouse by removing an arbitrary integer amount of cubic meters of hay from the top of each stack. You can take different amounts of hay from different haystacks. Besides, it is allowed not to touch a stack at all, or, on the contrary, to remove it completely. If a stack is completely removed, the corresponding cell becomes empty and no longer contains the stack.

Polycarp wants the following requirements to hold after the reorganization:

the total amount of hay remaining in the warehouse must be equal to k, the heights of all stacks (i.e., cells containing a non-zero amount of hay) should be the same, the height of at least one stack must remain the same as it was, for the stability of the remaining structure all the stacks should form one connected region. The two stacks are considered adjacent if they share a side in the table. The area is called connected if from any of the stack in the area you can get to any other stack in this area, moving only to adjacent stacks. In this case two adjacent stacks necessarily belong to the same area.

Help Polycarp complete this challenging task or inform that it is impossible.

Input

The first line of the input contains three integers n, m (1 ≤ n, m ≤ 1000) and k (1 ≤ k ≤ 1018) — the number of rows and columns of the rectangular table where heaps of hay are lain and the required total number cubic meters of hay after the reorganization.

Then n lines follow, each containing m positive integers ai, j (1 ≤ ai, j ≤ 109), where ai, j is equal to the number of cubic meters of hay making the hay stack on the i-th row and j-th column of the table.

Output

In the first line print "YES" (without quotes), if Polycarpus can perform the reorganisation and "NO" (without quotes) otherwise. If the answer is "YES" (without quotes), then in next n lines print m numbers — the heights of the remaining hay stacks. All the remaining non-zero values should be equal, represent a connected area and at least one of these values shouldn't be altered.

If there are multiple answers, print any of them.

Sample Input

2 3 35 10 4 9 9 9 7

Sample Output

YES 7 0 7 7 7 7

Hint

題意

給你一個n*m的矩陣，然後給你一個k

這個矩陣裡面的數，只能減小，不能增加。

然後你要是的矩陣最後只剩下一個連通塊，且連通塊裡面有一個位置的數沒有改變。

連通塊的權值和恰好等於k

讓你輸出一個解。

題解：

把所有數，從大到小排序，然後用並查集去維護

只要當前這個連通塊的大小大於等於k/a[i][j]就好了

然後輸出的時候用bfs去輸出，去維護這個連通塊的大小

然後就完了……

程式碼

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

const int maxn = 1e6+7;
int a[1003][1003],p[maxn],num[maxn],n,m,vis[1003][1003];
long long k;
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int fi(int x)
{
    return p[x]==x?x:p[x]=fi(p[x]);
}
void uni(int x,int y)
{
    int p1=fi(x),p2=fi(y);
    if(p1==p2)return;
    p[p1]=p2;
    num[p2]+=num[p1];
    num[p1]=0;
}
struct node
{
    int x,y,z;
    node(int x1,int y1,int z1){x=x1,y=y1,z=z1;}
};
bool cmp(node a,node b)
{
    return a.x>b.x;
}
vector<node>Q;
void solve(int x,int y,int z,int v)
{
    queue<node>Q;
    Q.push(node(x,y,0));
    vis[x][y]=1;z--;
    while(!Q.empty())
    {
        node now = Q.front();
        Q.pop();
        for(int i=0;i<4;i++)
        {
            int xx=now.x+dx[i];
            int yy=now.y+dy[i];
            if(z==0)continue;
            if(xx<=0||xx>n)continue;
            if(yy<=0||yy>m)continue;
            if(vis[xx][yy])continue;
            if(a[xx][yy]<v)continue;
            vis[xx][yy]=1,z--;
            Q.push(node(xx,yy,0));
        }
    }
    for(int i=1;i<=n;i++,cout<<endl)
        for(int j=1;j<=m;j++)
            if(vis[i][j])printf("%d ",v);
            else printf("0 ");
}
int main()
{
    scanf("%d%d%lld",&n,&m,&k);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            scanf("%d",&a[i][j]),Q.push_back(node(a[i][j],i,j));
    for(int i=0;i<maxn;i++)p[i]=i,num[i]=1;
    sort(Q.begin(),Q.end(),cmp);
    for(int i=0;i<Q.size();i++)
    {
        int v = Q[i].x;if(v==0)break;
        int x = Q[i].y;
        int y = Q[i].z;
        for(int j=0;j<4;j++)
        {
            int xx = x+dx[j];
            int yy = y+dy[j];
            if(a[xx][yy]>=v)uni((xx-1)*m+yy,(x-1)*m+y);
        }
        long long Num = k/v;
        if(k%v)continue;
        int fa=fi((x-1)*m+y);
        if(Num<=num[fa])
        {
            printf("YES\n");
            solve(x,y,Num,v);
            return 0;
        }
    }
    printf("NO\n");
}

這個題目就是把無序化的bfs變為 有序化的。怎麼有序化？

可以發現，列舉一個節點的值為基準時，他的聯通部分權值比他小的點是無用的！所以只需知道比他權值大的且和他聯通的點有多少個就行了。

如果把所有點按權值由高到低排序，那麼我們就可以用並查集維護了。每次列舉一個點時，判斷與他相鄰的點是否比他大，如果比他大就並查集unite一下。

很不錯的題目。

最後注意一下並查集的初始化和getid的計算。如果getid充分利用了(0,m*n)的話，初始化還好。如果沒有充分利用，初始化就應該考慮到多出來的一些點！

#include<bits/stdc++.h>
#define rep(i, j, k) for (int i=j; i<k; i++)
#define ll long long 
#define dprintf if (debug) printf
using namespace std; 
const int maxn = 1005;
const int debug = 0;
int n, m, cnt;
ll k, a[maxn][maxn];
int vis[maxn][maxn], par[maxn*maxn],high[maxn*maxn];
ll num[maxn*maxn];
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};
struct Node{
	int x, y;
	ll w;
}st[maxn * maxn], ans[maxn * maxn];

queue<Node> Q;
int findfather(int x)
{
    return par[x]=par[x]==x?x:findfather(par[x]);
}
 
bool unite(int a, int b)
{
    int fa=findfather(a), fb=findfather(b);
    if(fa==fb)
        return false;
    if(high[fa]>high[fb]){
        par[fb]=fa;
        num[fa] += num[fb];
    }
    else
    {
        par[fa]=fb;
    	num[fb] += num[fa];
        if(high[fa]==high[fb])
            ++high[fb];
    }
    return true;
}

//初始化



int bfs(Node now){
	ll goal = k/a[now.x][now.y]; ll val = a[now.x][now.y];
	if (goal > m*n) return 0;
	ll cnt = 1;
	ans[0] = {now.x, now.y};
	while (!Q.empty()) Q.pop();
	memset(vis, 0, sizeof(vis));
	Q.push(now); vis[now.x][now.y] = 1;
	while (!Q.empty()){
		if (cnt == goal) break;
	    now = Q.front(); Q.pop();
		dprintf("bfs %d %d\n", now.x, now.y);
	    int x = now.x; int y = now.y;
		rep(i, 0, 4){
			int nx = x + dx[i]; int ny = y + dy[i];
			if (nx<=0 || nx>n || ny<=0 || ny>m || a[nx][ny] < val || vis[nx][ny]) continue;
			vis[nx][ny] = 1;
			Q.push({nx, ny, a[nx][ny]});
			ans[cnt++] = {nx, ny};
			if (cnt == goal) break;
		}
	}
	//dprintf("cnt = %d\n", cnt);
	if (cnt == goal){
		rep(i, 0, cnt){
			int x = ans[i].x; int y = ans[i].y;
			a[x][y] = -1;
		}
		puts("YES");
		for (int i=1; i<=n; i++){
			for (int j=1; j<=m; j++){
				if (a[i][j] != -1)
					printf("0 ");
				else
					printf("%lld ", val);
			}
			puts("");
		}
		return 1;
	}
	return 0;
}

bool cmp(Node n1, Node n2){
	return n1.w > n2.w;
}
int id(int x, int y){///!!!
	return (x-1) * m + y-1;
}
int main(){
	scanf("%d%d%lld", &n, &m, &k);
	for (int i=1; i<=n; i++){
		for (int j=1; j<=m; j++){
			scanf("%lld", &a[i][j]);
			st[cnt++] = {i, j, a[i][j]};
		}
	}
	rep(i, 0, cnt){
		par[i] = i;
		num[i] = 1;
	}
	sort(st, st+cnt, cmp);
	rep(i, 0, cnt){
		int x = st[i].x; int y = st[i].y; ll w = st[i].w;
		dprintf("\n\n x = %d y = %d w = %lld\n", x, y, st[i].w);
		rep(j, 0, 4){
			int nx = x + dx[j]; int ny = y + dy[j];
			if (nx <=0 || nx > n || ny <=0 || ny>m) continue;
			if (a[nx][ny] >= w) {
				dprintf("unite %d %d %d %d\n", x, y, nx, ny);
				
				unite(id(x, y), id(nx, ny));
				dprintf("numfa = %lld\n", num[findfather(id(x, y))]);
			}
		}
		if (k%w) continue;
		if (num[findfather(id(x, y))] >= k/w){
			bfs(st[i]);
			return 0;
		}
		//dprintf("k = %lld, w = %lld, k/w = %lld\n", k, w, k/w);
	}
	puts("NO");
	return 0;
}