luogu P6787 「SWTR-6」Snow Mountain

题面传送门
月赛中的简单题。
考虑将原数列以能量排序，那么肯定是前一半匹配后一半最优。
而匹配时优先取被覆盖次数最大的即可判断无解。
但是有一种特殊情况，就是前一半全部指向一个点。
那么把这个点特殊处理掉然后正常做即可。
代码实现:

#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m,k,x,y,z,in[1000039],f1[1000039],f2[1000039],tot,place,xs[1000039],flag[1000039],flags,head1,head2;
long long ans;
struct yyy{
    int num,x,y;
    bool operator <(const yyy &y)const{
    	return x<y.x;
	}  
}tmp1,tmp2,tmp3;
priority_queue<yyy> q;
struct ques{int x,y,num;}s[500039];
inline bool cmp(ques x,ques y){return x.x<y.x;}
int main(){
//	freopen("1.in","r",stdin);
	register int i;
	scanf("%d",&n);
	for(i=1;i<=n;i++) scanf("%d",&s[i].x),xs[i]=s[i].x;
	for(i=1;i<=n;i++) scanf("%d",&s[i].y),s[i].num=i;
	sort(s+1,s+n+1,cmp);
	for(i=1;i<=n/2;i++) if(s[i].y!=-1) in[s[i].y]++;
	for(i=n/2+1;i<=n;i++)  tot=max(tot,in[s[i].num]);
	if(tot==n/2){
		for(i=n/2+1;i<=n;i++)if(in[s[i].num]==n/2){place=i;break;}
		for(i=n/2+1;i<=n;i++) if(place!=i&&s[place].y!=s[i].num&&s[i].y!=s[place].num){ans+=min(s[place].x,s[i].x);flag[i]=flag[place]=1;f1[1]=s[place].num;f2[1]=s[i].num;break;}
		if(!ans) {printf("-1\n");return 0;}
	    head1=n/2-1;head2=n/2;
		for(i=2;i<=n/2;i++) {
			while(flag[head1]) head1--;
			while(flag[head2]) head2++;
	    	ans+=(long long)s[head1].x*i;
	    	f1[i]=s[head1].num;f2[i]=s[head2].num;
	    	flag[head1]=flag[head2]=1;
		}
		printf("%lld\n",ans);
		for(i=1;i<=n/2;i++) printf("%d %d\n",f1[i],f2[i]);
	    return 0;
	}
	//for(i=1;i<=n;i++) printf("%d\n",s[i].y);
	for(i=n/2+1;i<=n;i++)q.push((yyy){s[i].num,in[s[i].num],s[i].y});
	for(i=n/2;i>=1;i--){
		tmp1=q.top();
		q.pop();
		if(tmp1.num==s[i].y) {
			if(q.empty()){printf("-1\n");return 0;}
			tmp2=q.top();
			q.pop();
			q.push(tmp1);
			f1[n/2-i+1]=s[i].num;f2[n/2-i+1]=tmp2.num;
			ans+=(long long)(n/2-i+1)*s[i].x;
		}
		else f1[n/2-i+1]=s[i].num,f2[n/2-i+1]=tmp1.num,ans+=(long long)(n/2-i+1)*s[i].x;
	} 
	printf("%lld\n",ans);
	for(i=1;i<=n/2;i++) printf("%d %d\n",f1[i],f2[i]);
}