CSUOJ 1555 Inversion Sequence 线段树/STL

Inversion Sequence

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%lld & %llu

 

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

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

typedef long long LL;

const int maxn=10005;

int sum[maxn<<2];
int pos,ok;
int n;
int k;
int ans[maxn];

void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}

void build(int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]=1;
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}

void query(int t,int l,int r,int rt)
{
    if(t>sum[rt])
    {
        ok=0;
        return ;
    }
    if(l==r)
    {
        pos=l;
        sum[rt]=0;
        return ;
    }
    int m=(l+r)>>1;
    if(t<=sum[rt<<1])
        query(t,lson);
    else
        query(t-sum[rt<<1],rson);

        pushup(rt);
}

int main()
{
    //freopen("input.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        ok=1;
        build(1,n,1);
        for(int i=1; i<=n; i++)
            {
                scanf("%d",&k);
                if(!ok)
                    continue;
                query(k+1,1,n,1);
                ans[pos]=i;
            }

        if(!ok)
        {
            printf("No solution
");
            continue;
        }
        printf("%d",ans[1]);
        for(int i=2;i<=n;i++)
            printf(" %d",ans[i]);
        printf("
");

    }
}

　　

第二种方法是用vector数组来做
考虑到它能够插♂来♂插♂去♂的属性

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

typedef long long LL;

const int maxn=10005;

int a[maxn];

vector<int> vec;

int main()
{
    //freopen("input.txt","r",stdin);
    int n;
    int k;
    while(~scanf("%d",&n))
    {
        vec.clear();
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);

        int flag=1;
        for(int i=n; i>=1; i--)
        {
            if(vec.size()<a[i])
            {
                flag=0;
                break;
            }
            vec.insert(vec.begin()+a[i],i);
        }
        if(flag)
        {
            for(int i=0; i<vec.size(); i++)
            {
                if(i!=0)
                    printf(" ");
                printf("%d",vec[i]);
            }
            printf("
");
        }
        else
            printf("No solution
");
    }
}