CodeForces - 1102B：Array K-Coloring（给数字涂颜色）

B. Array K-Coloring
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given an array a consisting of n integer numbers.

You have to color this array in k colors in such a way that:

Each element of the array should be colored in some color;
For each i from 1 to k there should be at least one element colored in the i-th color in the array;
For each i from 1 to k all elements colored in the i-th color should be distinct.
Obviously, such coloring might be impossible. In this case, print “NO”. Otherwise print “YES” and any coloring (i.e. numbers c1,c2,…cn, where 1≤ci≤k and ci is the color of the i-th element of the given array) satisfying the conditions above. If there are multiple answers, you can print any.

Input
The first line of the input contains two integers n and k (1≤k≤n≤5000) — the length of the array a and the number of colors, respectively.

The second line of the input contains n integers a1,a2,…,an (1≤ai≤5000) — elements of the array a.

Output
If there is no answer, print “NO”. Otherwise print “YES” and any coloring (i.e. numbers c1,c2,…cn, where 1≤ci≤k and ci is the color of the i-th element of the given array) satisfying the conditions described in the problem statement. If there are multiple answers, you can print any.

Examples
inputCopy
4 2
1 2 2 3
outputCopy
YES
1 1 2 2
inputCopy
5 2
3 2 1 2 3
outputCopy
YES
2 1 1 2 1
inputCopy
5 2
2 1 1 2 1
outputCopy
NO
Note
In the first example the answer 2 1 2 1 is also acceptable.

In the second example the answer 1 1 1 2 2 is also acceptable.

There exist other acceptable answers for both examples.
题意：给n个数字、k种颜色，要求你给每个数字涂色，但是每种颜色中数字不能相同，并且颜色都要用完，当满足条件时，就输出YES以及涂的颜色，否则输出NO
思路：
1、题目要求每种颜色的数字不能相同，那么我们就可以先判断，如果一个数字出现的次数已经大于颜色的种类，那么必然不能满足条件
2、然后我们再满足条件颜色都要用完这个条件，只要n>k那么上面这个条件就可以满足
3、我们可能需要一个二维数组来记录每种颜色中的数字都有哪些，通过这个数组，我们就可以让接下来n-k个数字完成涂色
代码如下

#include 
#include 
#include 
#include 
#include 
#define MAX_SIZE 5010
using namespace std;

//a用来存储数字，sum记录数字出现次数，ans记录每个数字所涂的颜色
int a[MAX_SIZE],sum[MAX_SIZE],ans[MAX_SIZE];
bool vis[MAX_SIZE][MAX_SIZE];	//这里使用bool的数组为了节省空间

int main()
{
    int i,j,k,n,temp;
    scanf("%d %d",&n,&k);
    for(i = 0;i < n;i++)
        scanf("%d",&a[i]);
    //判断一个数字出现的次数
    for(i = 0;i < n;i++)
        sum[a[i]]++;
    for(i = 0;i <= 5000;i++)
    {
        if(sum[i] > k)
        {
            printf("NO\n");
            return 0;
        }
    }
    //判断数字个数和颜色种类
    if(k > n)
    {
        printf("NO\n");
        return 0;
    }
    
    int pos = 1;
    for(i = 0;i < n;i++)
    {
        //先将每种颜色都使用
        if(pos <= k)
        {
            ans[i] = pos;
            vis[pos][a[i]] = 1;
            pos++;
        }
        else
        {
            //之后寻找那种颜色没有数字a[i]，那么就将数字a[i]涂为j的颜色
            for(j = 1;j <= k;j++)
            {
                if(vis[j][a[i]] == 0)
                {
                    ans[i] = j;
                    vis[j][a[i]] = 1;
                    break;
                }
            }
        }
    }
    printf("YES\n");
    for(i = 0;i < n;i++)
        printf("%d ",ans[i]);
    return 0;
}