Codeforces Global Round 14, C. Phoenix and Towers

problem

C. Phoenix and Towers
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Phoenix has n blocks of height h1,h2,…,hn, and all hi don’t exceed some value x. He plans to stack all n blocks into m separate towers. The height of a tower is simply the sum of the heights of its blocks. For the towers to look beautiful, no two towers may have a height difference of strictly more than x.

Please help Phoenix build m towers that look beautiful. Each tower must have at least one block and all blocks must be used.

Input
The input consists of multiple test cases. The first line contains an integer t (1≤t≤1000) — the number of test cases.

The first line of each test case contains three integers n, m, and x (1≤m≤n≤105; 1≤x≤104) — the number of blocks, the number of towers to build, and the maximum acceptable height difference of any two towers, respectively.

The second line of each test case contains n space-separated integers (1≤hi≤x) — the heights of the blocks.

It is guaranteed that the sum of n over all the test cases will not exceed 105.

Output
For each test case, if Phoenix cannot build m towers that look beautiful, print NO. Otherwise, print YES, followed by n integers y1,y2,…,yn, where yi (1≤yi≤m) is the index of the tower that the i-th block is placed in.

If there are multiple solutions, print any of them.

Example
inputCopy
2
5 2 3
1 2 3 1 2
4 3 3
1 1 2 3
outputCopy
YES
1 1 1 2 2
YES
1 2 2 3
Note
In the first test case, the first tower has height 1+2+3=6 and the second tower has height 1+2=3. Their difference is 6−3=3 which doesn’t exceed x=3, so the towers are beautiful.

In the second test case, the first tower has height 1, the second tower has height 1+2=3, and the third tower has height 3. The maximum height difference of any two towers is 3−1=2 which doesn’t exceed x=3, so the towers are beautiful.

solution

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
struct node{int v, id; bool operator < (node b)const{return v>b.v;}
};
int h[maxn], cc[maxn];
int main(){int T;  cin>>T;while(T--){//memset(h,0,sizeof(h));int n, m, x;  cin>>n>>m>>x;priority_queue<node>q;for(int i = 1; i <= n; i++){int t;  cin>>t;  q.push({t,i});h[i] = 0;}int be = 1;while(q.size()){node t = q.top(); q.pop();h[be] += t.v;cc[t.id] = be;be++;if(be==m+1)be = 1;}int mx = 0, mi = 1e9+10;for(int i = 1; i <= m; i++){mx = max(mx, h[i]);mi = min(mi, h[i]);}if(mx-mi > x)cout<<"NO\n";else {cout<<"YES\n";for(int i = 1; i <= n; i++)cout<<cc[i]<<" ";cout<<"\n";}}return 0;
}