CodeForces - 48D Permutations题解

xiaoxiao2021-02-28  46

A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. For example, (1)(4, 3, 5, 1, 2)(3, 2, 1) are permutations, and (1, 1)(4, 3, 1)(2, 3, 4) are not.

There are many tasks on permutations. Today you are going to solve one of them. Let’s imagine that somebody took several permutations (perhaps, with a different number of elements), wrote them down consecutively as one array and then shuffled the resulting array. The task is to restore the initial permutations if it is possible.


The first line contains an integer n (1 ≤ n ≤ 105). The next line contains the mixed array of n integers, divided with a single space. The numbers in the array are from 1 to 105.


If this array can be split into several permutations so that every element of the array belongs to exactly one permutation, print in the first line the number of permutations. The second line should contain n numbers, corresponding to the elements of the given array. If the i-th element belongs to the first permutation, the i-th number should be 1, if it belongs to the second one, then its number should be 2 and so on. The order of the permutations’ numbering is free.

If several solutions are possible, print any one of them. If there’s no solution, print in the first line  - 1.

Examples Input 9 1 2 3 1 2 1 4 2 5 Output 3 3 1 2 1 2 2 2 3 2 Input 4 4 3 2 1 Output 1 1 1 1 1 Input 4 1 2 2 3 Output -1





#include <cstdio> #include <algorithm> #include <cstring> #include <vector> #include <map> using namespace std; const int MAX_N=1e5+10; const int INF = 0x3f3f3f3f; int n; int b[MAX_N]; bool used[MAX_N]; int p; struct node{ int value; int id; }a[MAX_N]; int cmp(node x,node y){ return x.value<y.value; } bool dfs(node num,int i){ used[i]=true; if(num.value==1){ return true; } for(int j=i-1;j>=0;j--){ //找不到 if(j==0 && a[j].value!=num.value-1){ return false; } if(!used[j] && a[j].value==num.value-1){ used[j]=true; b[a[j].id]=p; if(dfs(a[j],j)){ return true; }else{ return false; } } } return false; } int main(){ while(~scanf("%d",&n)){ memset(used,false,sizeof(used)); memset(b,0,sizeof(b)); for(int i=0;i<n;i++){ int tmp; scanf("%d",&tmp); a[i].id=i; a[i].value=tmp; } sort(a,a+n,cmp); p=1; bool flag=true; for(int i=n-1;i>=0;i--){ if(!used[i]){ used[i]=true; b[a[i].id]=p; //找不到 if(!dfs(a[i],i)){ flag=false; break; } p++; } } if(flag){ printf("%d\n",p-1); for(int i=0;i<n;i++){ if(i==0){ printf("%d",b[i]); }else{ printf(" %d",b[i]); } } printf("\n"); }else{ printf("-1\n"); } } return 0; }