Codeforces 659F Polycarp and Hay (排序+並查集)

解析：將所有的cell按高度從大到小排序，然後每次放入值相同的cell，用並查集維護集合和集合大小，並判斷是否滿足條件即可。找到後搜索標記所有的位置。

[code]:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<map>

#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define lowbit(i) (i&-i)
using namespace std;
typedef long long LL;

struct Node{
    int id,val;
    bool operator < (const Node &z) const{
        return val > z.val;
    }
}C[1005*1005];
int n,m,a[1005][1005],b[1005][1005];
LL total,num[1005*1005];
int par[1005*1005],vis[1005*1005];
int dx[4] = {0,0,-1,1};
int dy[4] = {-1,1,0,0};

void init(){
    for(int i = 0;i < n*m;i++){
        par[i] = i;
        num[i] = 1;
    }
}
int find(int x){
    return par[x]==x?x:par[x]=find(par[x]);
}
void unite(int x,int y){
    x = find(x),y = find(y);
    if(x == y) return;
    par[y] = x;
    num[x] += num[y];
}
LL check(int x,int y){
    int k,nx,ny;
    for(k = 0;k < 4;k++){
        nx = x +dx[k];
        ny = y +dy[k];
        if(nx<0||nx>=n||ny<0||ny>=m||!vis[nx*m+ny]) continue;
        unite(x*m+y,nx*m+ny);
    }
    return num[find(x*m+y)];
}

void dfs(int x,int y,int val){
    if(vis[x*m+y]==0) return;
    if(total > 0){
        b[x][y] = val;
        total -= val;
    }
    vis[x*m+y] = 0;
    int k,nx,ny;
    for(k = 0;k < 4;k++){
        nx = x +dx[k];
        ny = y +dy[k];
        if(nx<0||nx>=n||ny<0||ny>=m) continue;
        dfs(nx,ny,val);
    }
}

int main(){
    int i,j;
    scanf("%d%d%I64d",&n,&m,&total);
    init();
    for(i = 0;i < n;i++){
        for(j = 0;j < m;j++){
            scanf("%d",&a[i][j]);
            C[i*m+j].id = i*m+j;
            C[i*m+j].val = a[i][j];
        }
    }
    sort(C,C+n*m);
    LL tmp,cnt;bool flag,ok;
    for(i = 0;i < n*m;i = j){
        tmp = total/C[i].val;
        flag = total%C[i].val==0;
        ok = 0;
        for(j = i;j < n*m&&C[j].val==C[i].val;j++){
            vis[C[j].id] = 1;
            cnt = check(C[j].id/m,C[j].id%m);
            if(flag&&cnt>=tmp){
                ok = 1;
                break;
            }
        }
        if(ok){
            puts("YES");
            dfs(C[j].id/m,C[j].id%m,C[j].val);
            int ii,jj;
            for(ii = 0;ii < n;ii++){
                for(jj = 0;jj < m;jj++){
                    printf("%d ",b[ii][jj]);
                }
                putchar('\n');
            }
            return 0;
        }
    }
    puts("NO");

    return 0;
}