2012 Asia Changsha Regional Contest Problem G

Graph Reconstruction

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Time Limit: 2 Seconds                                     Memory Limit: 65536 KB                                                     Special Judge                            

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Let there be a simple graph with N vertices but we just know the degree of each vertex. Is it possible to reconstruct the graph only by these information? 

A simple graph is an undirected graph that has no loops (edges connected at both ends to the same vertex) and no more than one edge between any two different vertices. The degree of a vertex is the number of edges that connect to it.

Input

There are multiple cases. Each case contains two lines. The first line contains one integer N (2 ≤ N ≤ 100), the number of vertices in the graph. The second line conrains N integers in which the i_th item is the degree of i_th vertex and each degree is between 0 and N-1(inclusive).

Output

If the graph can be uniquely determined by the vertex degree information, output "UNIQUE" in the first line. Then output the graph.

If there are two or more different graphs can induce the same degree for all vertices, output "MULTIPLE" in the first line. Then output two different graphs in the following lines to proof.

If the vertex degree sequence cannot deduced any graph, just output "IMPOSSIBLE".

The output format of graph is as follows:

N E
u

₁

 u

₂

 ... u

_E

v

₁

 v

₂

 ... v

_EWhere N is the number of vertices and E is the number of edges, and {u_i,v_i} is the i_th edge the the graph. The order of edges and the order of vertices in the edge representation is not important since we would use special judge to verify your answer. The number of each vertex is labeled from 1 to N. See sample output for more detail.

Sample Input

1
0
6
5 5 5 4 4 3
6
5 4 4 4 4 3
6
3 4 3 1 2 0

Sample Output

UNIQUE
1 0


UNIQUE
6 13
3 3 3 3 3 2 2 2 2 1 1 1 5
2 1 5 4 6 1 5 4 6 5 4 6 4
MULTIPLE
6 12
1 1 1 1 1 5 5 5 6 6 2 2
5 4 3 2 6 4 3 2 4 3 4 3
6 12
1 1 1 1 1 5 5 5 6 6 3 3
5 4 3 2 6 4 3 2 4 2 4 2
IMPOSSIBLE

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                            Author: WANG, Yelei                                                     Contest: The 2013 ACM-ICPC Asia Changsha Regional Contest

#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff
#define mod 1000000007
#define ll long long
#define maxn 100005
#define pi acos(-1.0)
int n, m, k, t, flag, ans, s ,x ,y,c;
bool cmp(pair<int,int> a, pair<int,int> b){ return a.first > b.first; }
pair<int, int>a[maxn],b[maxn],pa[maxn],pb[maxn];
int main(){
    while (~scanf("%d\n",&n)){
        flag = 1; s = 0; k = 0, t = 0; c = 0;
        for (int i = 1; i <= n; i++){
            cin >> a[i].first;
            if (a[i].first > n || a[i].first < 0)flag = 0;
            s += a[i].first;
            if (a[i].first == 0)t++;
            a[i].second = i;
        }
        if (t == n){ printf("UNIQUE\n"); printf("%d %d\n\n\n", n, 0); continue; }
        if ((s & 1)||flag==0){ printf("IMPOSSIBLE\n"); continue; }
        sort(a+1, a + n+1, cmp);
        for (int i = 1; i <= n; i++)b[i].first = a[i].first, b[i].second = a[i].second;
        for (int i = 1, j = 1; i <= n; i++){
            sort(a + i , a + n + 1, cmp);
            m = a[i].first;
            if (m < 0||m+i>n){ flag = 0; break; }
            for (j = i + 1; j <= i + m; j++){
                a[j].first--;
                pa[k++] = make_pair(a[i].second, a[j].second);
            }
        }
        for (int i = 1, j = 1; i <= n; i++){
            sort(b + i , b + n + 1, cmp);
            m = b[i].first;
            if (b[i+m].first != 0 && i+m < n&&b[i+m].first == b[i+m+1].first)swap(b[i+m].second, b[i+m+1].second), flag = 2;
            for (j = i + 1; j <= i + m; j++){
                pb[c++] = make_pair(b[i].second, b[j].second);
                b[j].first--;
            }
        }
        if (flag == 0||a[n].first!=0){ printf("IMPOSSIBLE\n"); continue; }
        if (flag == 1)printf("UNIQUE\n"),printf("%d %d\n",n,k);
        if (flag == 2)printf("MULTIPLE\n"),printf("%d %d\n",n,k);
        for (int i = 0; i < k; i++)printf(i == k - 1 ? "%d\n" : "%d ", pa[i].first);
        for (int i = 0; i < k; i++)printf(i == k - 1 ? "%d\n" : "%d ", pa[i].second);
        if (flag == 2){
            printf("%d %d\n", n, k);
            for (int i = 0; i < k; i++)printf(i == k - 1 ? "%d\n" : "%d ", pb[i].first);
            for (int i = 0; i < k; i++)printf(i == k - 1 ? "%d\n" : "%d ", pb[i].second);
        }
    }
    return 0;
}