1283C 构造

对于已经将礼物送出去的i，即f[i]>0的，不用改变也不能改变，只需把剩下的没发礼物的和没收到礼物的，让他们一一对应，并且不出现f[i]=i的情况，也就是没有人将礼物发给自己
vector f(n + 1), hasGift(n + 1, 0);
vector senders, receivers;
较难点在于这个对应情况怎么处理，对于send和receive，如果出现值相同，让swap(receivers[i],receivers[(i+1)%len]);

include <bits/stdc++.h>

using namespace std;

define int long long

const int N=5010;
vector g[N];
//int a[N];
int dx[]={-1,1,0,0};
int dy[]={0,0,-1,1};
void solve()
{
int n,i,j;
cin >> n;
vector f(n+1),hasgift(n+1);
vector send,receiver;
for(i=1;i<=n;i++)
{
cin >> f[i];
if(f[i])
hasgift[f[i]]=i;
else send.push_back(i);
}
for(i=1;i<=n;i++)
{
if(!hasgift[i])
receiver.push_back(i);
}
int len=send.size();
for(i=0;i<send.size();i++)
{
if(send[i]==receiver[i])
swap(receiver[i],receiver[(i+1)%len]);
}
for(i=0;i<len;i++)
f[send[i]]=receiver[i];
for(i=1;i<=n;i++)
cout << f[i] << " ";
cout <<endl;
return ;
}
signed main ()
{
// int t;
//cin >> t;
//while(t--)
solve();
return 0;
}