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Average

3
6
1 0 1 0 0 0
5
1 1 1 1 1
3
1 2 3

NO
YES
0
YES
2
2 1
3 2

首先一个人有三种操作，每个人选择的操作影响周边2人.

如果无环的化可直接贪心。

所以枚举人1和人2的关系，把环从1与2之间断开

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define mp make_pair
#define MAXN (100000+10)
typedef long long ll;
int n;
int a[MAXN],a2[MAXN];
void rel(int &a,int &b,int c)
{
	a=c-1,b=c+1;
	if (a==0) a=n;if (b>n) b=1;
}
pair<int,int>ans[MAXN];
int siz=0;
bool check(int a[])
{
	a[n+1]=a[1];
	Fork(i,2,n)
	{
		if (a[i]==-1) a[i]++,a[i+1]--,ans[siz++]=mp(i+1<=n?i+1:1,i);
		else if (a[i]==1) a[i]--,a[i+1]++,ans[siz++]=mp(i,i+1<=n?i+1:1);
	}
	a[1]=a[n+1];
	For(i,n) if (a[i]) return 0;
	return 1;
}
void pri()
{
	cout<<"YES\n"<<siz<<'\n';
	Rep(i,siz) printf("%d %d\n",ans[i].first,ans[i].second);
}
int main()
{
//	freopen("A.in","r",stdin);
	
	int T;
	cin>>T;
	For(kcase , T)
	{
		cin>>n;
		For(i,n) scanf("%d",&a[i]);
		ll sum=0;
		For(i,n) sum+=a[i];
		if (sum%n){
			cout<<"NO\n";continue;
		}
		sum/=n;
		For(i,n) a[i]-=sum;
		
		bool flag=0;
		For(i,n) if (a[i]<-2||a[i]>2){
			cout<<"NO\n"; flag=1;break;
		}
		if (flag==1) continue;
		
		
		
		
		memcpy(a2,a,sizeof(a));
		MEM(ans) siz=0;
		if (check(a2)) 
		{
			pri();
		 	continue;
		} 

		memcpy(a2,a,sizeof(a));
		MEM(ans) siz=0;
		a2[1]--;a2[2]++;ans[siz++]=mp(1,2);
		if (check(a2)) 
		{
			pri();
		 	continue;
		} 
		
		memcpy(a2,a,sizeof(a));
		MEM(ans) siz=0;
		a2[1]++;a2[2]--;ans[siz++]=mp(2,1);
		if (check(a2)) 
		{
			pri();
		 	continue;
		} 
		
		
		
		cout<<"NO\n";
	}
	
	return 0;
}

HDU 5353(Average-贪心分果)

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原文地址：http://blog.csdn.net/nike0good/article/details/47430367