1283C - Friends and Gifts（思维）

传送门

思路：开两个数组，分别保存没送出礼物的人和没收到礼物的人。然后遍历两个数组，发现两个数组元素有相等的地方则与下一个交换，注意边界当最后一个相同则和开始的元素交换。

Code:

#include<iostream>
#include<string>
#include<map>
#include<algorithm>
#include<memory.h>
#include<cmath>
#include<queue>
#include<sstream>
#define pii pair<int,int>
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const int Max = 6e6 + 5;
int lst[Max];
int b1[Max],b2[Max];
int ls1[Max], ls2[Max];
int main()
{
   
	FAST;
	int n;cin >> n;
	map<int, int> ma;
	vector<int> vec;
	for (int i = 1;i <= n;i++)
	{
   
		cin >> lst[i];
		if (lst[i] != 0)b1[i]++;
		b2[lst[i]]++;
	}
	int g = 0, f = 0;
	for (int i = 1;i <= n;i++)
	{
   
		if (b1[i] == 0)ls1[++g] = i;
		if (b2[i] == 0)ls2[++f] = i;
	}
	for (int i = 1;i <= g;i++)
	{
   
		if (ls1[i] == ls2[i])
		{
   
			if (i == g)swap(ls2[g], ls2[1]);
			else swap(ls2[i], ls2[(i + 1)]);
		}
	}g = 0;
	for (int i = 1;i <= n;i++)
	{
   
		if (lst[i])cout << lst[i] << " ";
		else cout << ls2[++g] << " ";
	}
}