ACM: 强化训练-Inversion Sequence-线段树 or STL·vector

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

这题有两种做法，一开始直接用STL里的vector A掉了这个题目，然后基神让我试下用线段树来解这个题目，我就两种方法都写了

先来线段树： 从小到大插入到第k个格子。

 

/*/
线段树做法： 
首先，弄个长度为n的线段树，维护sum，初始值为1
那么，sum[1]等于n，代表着空着的位置的数量
现在要把某个数字插入到p位置，就在线段树找到一个位置x，使得[x]里面的sum等于p； 
然后把那个1修改为0，就表示那个位置不是空着的了。 


/*/ 

#include"iostream"
#include"cstdio"
#include"cstring"
#include"algorithm"
#include"cmath"
using namespace std;
#define MX 11111
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int sum[MX<<2];

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

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

int Query(int p,int l,int r,int rt) {
	if(l==r) {      	//找到点，插入，返回点的位置值 
		sum[rt]--;
		return l;
	}
	int m=(l+r)>>1,ret=0;
	if(p<=sum[rt<<1]) ret=Query(p,lson);//点在左边，直接向左找 
	else ret=Query(p-sum[rt<<1],rson);// 减去左边没去找的点
	PushUp(rt);
	return ret;
}
int a[MX];
int ans[MX];
int main() {
	int n;
	while(~scanf("%d",&n)) {
		Build(1,n,1);
		for(int i=1; i<=n; i++) {
			scanf("%d",&a[i]);
		}
		bool check=1;
		for(int i=1; i<=n; i++) {
			if(sum[1]<a[i]+1) {
				check=0;
				break;
			} else {
				int p= Query(a[i]+1,1,n,1);
				ans[p]=i;
			}
		}
		if(!check)printf("No solution\n");
		else {
			int first=1;
			for(int i=1; i<=n; i++) {
				if(first)first=0;
				else printf(" ");
				printf("%d",ans[i]);
			}
			printf("\n");
		}
	}
	return 0;
}

　　

 

然后是STL：  从大到小插入到第k个格子。

/*/
STL-vector:
从最大的数开始插入队列； 
利用vector中的insert()函数去将数字插入相应的位置；
这种方法暴力，简单。。。 
但是每一次插入复杂度是O(n);
数据有点放水，还是A了。
/*/ 
#include"iostream"
#include"cstdio"
#include"cstring"
#include"string"
#include"algorithm"
#include"vector"

using namespace std;
#define MX 10050
int num[MX];

int main() {
	int n;
	while(~scanf("%d",&n)) {
		for(int i=0; i<n; i++) {
			scanf("%d",&num[i]);
		}
		int flag=1;
		vector<int > v;
		v.push_back(0);
		for(int i=n-1;i>=0; i--) {
			if(v.size()<=num[i]) {
				printf("No solution\n");
				flag=0;
				break;
			}
			v.insert(v.begin()+num[i],i+1);
		}
		if(flag) {
			int first=1;
			vector<int>::iterator it;
			for(it=v.begin(); it!=v.end()-1; it++) {  
				if(first) {
					first=0;
					printf("%d",(*it));
				}
				else printf(" %d",(*it));
			}
			printf("\n");
		}
	}
	return 0;
}