CSUOJ 1555 Inversion Sequence

1555: Inversion Sequence

Time Limit: 2 Sec  Memory Limit: 256 MB
Submit: 107  Solved: 34

Description

For sequence i1, i2, i3, … , iN, we set aj to be the number of members in the sequence which are prior to j and greater to j at the same time. The sequence a1, a2, a3, … , aN is referred to as the inversion sequence of the original sequence (i1, i2, i3, … , iN). For example, sequence 1, 2, 0, 1, 0 is the inversion sequence of sequence 3, 1, 5, 2, 4. Your task is to find a full permutation of 1~N that is an original sequence of a given inversion sequence. If there is no permutation meets the conditions please output “No solution”.

Input

There are several test cases.
Each test case contains 1 positive integers N in the first line.(1 ≤ N ≤ 10000).
Followed in the next line is an inversion sequence a1, a2, a3, … , aN (0 ≤ aj < N)
The input will finish with the end of file.

Output

For each case, please output the permutation of 1~N in one line. If there is no permutation meets the conditions, please output “No solution”.

Sample Input

5
1 2 0 1 0
3
0 0 0
2
1 1

Sample Output

3 1 5 2 4
1 2 3
No solution

HINT

 

Source

解题：这个题目啊。。难读啊！

还是看例子吧

1 2 0 1 0

这个表示在1-N的排列中，存在这种排列，数字1前面只有1个数比他大，数字2前面只有2个比他大，数字3前面只有0个比他大，数字4前面只有1个比他大，数字5前面0个比他大。

所以答案 3 1 5 2 4

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 vector<int>v;
 int d[maxn],n;
 int main(){
     while(~scanf("%d",&n)){
         for(int i = ; i <= n; ++i)
             scanf("%d",d+i);
         v.clear();
         bool flag = true;
         for(int i = n; i > ; --i){
             if(v.size() < d[i]){
                 flag = false;
                 break;
             }
             v.insert(v.begin()+d[i],i);
         }
         if(flag){
             flag = false;
             for(int i = ; i < v.size(); ++i){
                 if(flag) putchar(' ');
                 printf("%d",v[i]);
                 flag = true;
             }
             putchar('\n');
         }else puts("No solution");
     }
     return ;
 }

线段树找空位置插入。。

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 struct node {
     int lt,rt,pos;
 } tree[maxn<<];
 int d[maxn],ans[maxn],n;
 bool flag;
 void build(int lt,int rt,int v) {
     tree[v].lt = lt;
     tree[v].rt = rt;
     if(lt == rt) {
         tree[v].pos = ;
         return;
     }
     int mid = (lt + rt)>>;
     build(lt,mid,v<<);
     build(mid+,rt,v<<|);
     tree[v].pos = tree[v<<].pos + tree[v<<|].pos;
 }
 void update(int s,int v,int value) {
     if(tree[v].lt == tree[v].rt) {
         tree[v].pos = ;
         ans[tree[v].lt] = value;
         return;
     }
     if(tree[v<<].pos >= s) update(s,v<<,value);
     else if(tree[v<<|].pos >= s - tree[v<<].pos) update(s-tree[v<<].pos,v<<|,value);
     else flag = false;
     tree[v].pos = tree[v<<].pos + tree[v<<|].pos;
 }
 int main() {
     while(~scanf("%d",&n)) {
         build(,n,);
         flag = true;
         for(int i = ; i <= n; ++i) {
             scanf("%d",d+i);
             if(flag) update(d[i]+,,i);
         }
         if(flag) for(int i = ; i <= n; ++i)
             printf("%d%c",ans[i],i == n?'\n':' ');
         else puts("No solution");
     }
     return ;
 }