Graph Reconstruction 可图化判定与构造方案

写了两小时发现这题没spj，绝望

#include<bits/stdc++.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string.h>
#include<iostream>
#include<math.h>
using namespace std;
#define ll long long

const int inf=0x3f3f3f3f;
#define FOR(n) for(int i=1;i<=n;i++)
#define pb push_back

const int maxn=1e2+24;
int n;

struct NODE{
    int id;
    int du;
    friend bool operator <(NODE ax,NODE bx){return ax.du>bx.du;}
}arr[maxn];

vector<int>U;
vector<int>V;
map<pair<int,int>,int>mp;
map<int,int>rmp;

int HavelHakimi(int n) //0不可图 1唯一图 -1多图 
{  
    int flag=0;
    for(int i = 1;i <= n; ++i){  
        sort(arr+i,arr+n+1);  
        if(arr[i].du + i > n) return 0;  
        /* 
        前面的i个顶点的度数已经足够了，现在剩余n-i个顶点， 
        现在从这n-i个顶点里面找出一个顶点它的度数为arr[i]， 
        arr[i]就代表与这个顶点相连的顶点个数，必然有arr[i]<n-i成立。 
        */  
        if(arr[i].du+i+1<=n && arr[arr[i].du+i].du==arr[arr[i].du+i+1].du){
            if(arr[arr[i].du+i].du!=0){
                flag=1;
                
                mp[make_pair(
                        arr[i].id,arr[arr[i].du+i].id)
                ]=arr[arr[i].du+i+1].id;

                rmp[arr[arr[i].du+i+1].id]=arr[arr[i].du+i].id;            
            }
        }
        for(int j = i+1;j <= arr[i].du+i;j++){  
            arr[j].du--;
            V.pb(arr[i].id);U.pb(arr[j].id);
            if(arr[j].du < 0) return 0;  
        }    
    }  
    if(arr[n].du != 0) return 0;  
    else if(flag)return -1;
    return 1; 
}

int main(){
    while(~scanf("%d",&n)){
        
        U.clear();V.clear();mp.clear();rmp.clear();

        int ducnt=0;
        FOR(n){
            scanf("%d",&arr[i].du);
            arr[i].id=i;
            ducnt+=arr[i].du;
        }
        int flag=HavelHakimi(n);
        //printf("%d
",flag);
        if(flag==0)printf("IMPOSSIBLE
");
        else if(flag==1){
            printf("UNIQUE
");
            printf("%d %d
",n,ducnt/2);
            for(int i=0;i<V.size();i++){
                printf("%d ",V[i]);
            }puts("");
            for(int i=0;i<U.size();i++){
                printf("%d ",U[i]);
            }
            puts("");
        }else{
            printf("MULTIPLE
");
            printf("%d %d
",n,ducnt/2);
            for(int i=0;i<V.size();i++){
                printf("%d ",V[i]);
            }puts("");
            for(int i=0;i<U.size();i++){
                printf("%d ",U[i]);
            }
            puts("");

            printf("%d %d
",n,ducnt/2);
            for(int i=0;i<V.size();i++){
                if(rmp.count(V[i])){
                    printf("%d ",rmp[V[i]]);
                    rmp.erase(V[i]);
                }
                else printf("%d ",V[i]);
            }
            puts("");
            for(int i=0;i<U.size();i++){
                if(mp.count(make_pair(V[i],U[i]))){
                    printf("%d ",mp[make_pair(V[i],U[i])]);
                }else if(rmp.count(U[i])){
                    printf("%d ",rmp[U[i]]);
                    rmp.erase(U[i]);
                }
                else printf("%d ",U[i]);
            }
            puts("");
        }
    }
}