给任意多个链表然后要合并成一个

 

 

 

 

 

 

 

 

 

 

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=110;
int pre[N],nex[N];
int z[N],y[N];
int wan[N];
int fa[N];
int findx(int x)
{
    return fa[x]=(x==fa[x]?x:findx(fa[x]));
}
int main()
{
    int n;
    scanf("%d",&n);
    int tot1=0,tot2=0,tot3=0;
    for(int i=0; i<=n; ++i) fa[i]=i;
    for(int  i=1; i<=n; ++i)
    {
        scanf("%d%d",&pre[i],&nex[i]);
        int x1=findx(i),x2=findx(pre[i]),x3=findx(nex[i]);
        if(x1!=x2&&x2!=0) fa[x2]=x1;
        if(x1!=x3&&x3!=0) fa[x3]=x1;
        if(pre[i]==0&&nex[i]==0)
        {
            wan[tot2++]=i;
            continue;
        }
        if(pre[i]==0) z[tot1++]=i;
        if(nex[i]==0) y[tot3++]=i;
    }
    bool used[120];
    memset(used,0,sizeof(used));
    for(int i=0; i<tot1-1; ++i)
    {
        int f1=findx(y[i]);
        for(int j=0; j<tot1; ++j)
        {
            int f2=findx(z[j]);
            if(f1!=f2&&!used[z[j]])
            {
                used[z[j]]=1;
                fa[f2]=f1;
                pre[z[j]]=y[i];
                nex[y[i]]=z[j];
                break;
            }
        }
    }
    int last,qi;
    if(tot2==n) last=1,qi=1;
    else last=y[tot1-1],qi=0;
    for(; qi<tot2; ++qi)
    {
        nex[last]=wan[qi];
        pre[wan[qi]]=last;
        last=wan[qi];
    }
    for(int i=1; i<=n; ++i)
        printf("%d %d\n",pre[i],nex[i]);
}
/*10
0 9
4 0
5 0
7 2
0 3
8 10
0 4
0 6
1 0
6 0*/