There are n players sitting at a round table. All of them have s cards
of n colors in total. Besides, initially the first person had cards of only the first color, the second one had cards of only the second
color and so on. They can swap the cards by the following rules:
as the players swap, a player can give a card of his color only;
a player can‘t accept a card of a color he already has (particularly, he can‘t take cards of his color, no matter whether he has given out all of them or not);
during one swap a pair of people swaps cards (each person gives one card and takes one card).
The aim of all n people is as follows: each of them should give out all the cards he had initially (that is, all cards of his color).
Your task is to denote whether such sequence of swaps is possible. If the answer is positive, you should list all the swaps.
Input
The first line contains integers n (1?≤?n?≤?200000)
and s (1?≤?s?≤?200000).
The second line contains n numbers, the i-th
number stands for how many cards the i-th player has by the moment the game starts. It is possible that a player has no cards initially.
Output
On the first line print "No" if such sequence of swaps is impossible. Otherwise, print "Yes".
If the answer is positive, next print number k — the number of the swaps. Then on k lines
describe the swaps by pairs of indices of the swapping players. Print the swaps and the numbers of the swaps in any order.
Sample test(s)
input
4 8
2 2 2 2
output
Yes
4
4 3
4 2
1 3
1 2
input
6 12
1 1 2 2 3 3
output
Yes
6
6 5
6 4
6 3
5 4
5 3
2 1
input
5 5
0 0 0 0 5
output
No
我们注意到：满足交换的基本条件是s%2==0&&a[i]<n
但是这还不能判断所有非法的。要交换我们很容易想到是首先
大的先交换，这样一来用优先队列搞下就好了。写的比较龊。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
const int INF=0x3f3f3f3f;
const int maxn=200000+100;
int n,s;
struct node{
int val;
int id;
bool operator<(node l1)const
{
return val<l1.val;
}
};
struct Node{
int id1;
int id2;
}ans[maxn];
node e[maxn];
int main()
{
int x;node st,ed;
priority_queue<node>q;
while(~scanf("%d%d",&n,&s))
{
while(!q.empty()) q.pop();
int flag=1;
REPF(i,1,n)
{
scanf("%d",&x);
if(x>=n)
flag=0;
st.val=x;
st.id=i;
if(st.val>0) q.push(st);
}
if(!flag||s%2)
{
puts("No");
continue;
}
int cnt=0;
for(int i=1;;i++)
{
if(q.empty()) break;
st=q.top();
q.pop();
if(q.size()<st.val)
{
flag=0;
break;
}
for(int j=0;j<st.val;j++)
{
ed=q.top();
q.pop();
e[j].val=ed.val;
e[j].id=ed.id;
e[j].val--;
ans[cnt].id1=st.id;
ans[cnt++].id2=e[j].id;
}
for(int j=0;j<st.val;j++)
{
if(e[j].val>0)
q.push(e[j]);
}
}
if(!flag)
{
puts("No");
continue;
}
puts("Yes");
printf("%d\n",s/2);
for(int i=0;i<cnt;i++)
printf("%d %d\n",ans[i].id1,ans[i].id2);
}
return 0;
}
