CF45-C. Dancing Lessons（STL或线段树）

C. Dancing Lessons

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n people taking dancing lessons. Every person is characterized by his/her dancing skill a_i. At the beginning of the lesson they line up from left to right. While there is at least one couple of a boy and a girl in the line, the following process is repeated: the boy and girl who stand next to each other, having the minimal difference in dancing skills start to dance. If there are several such couples, the one first from the left starts to dance. After a couple leaves to dance, the line closes again, i.e. as a result the line is always continuous. The difference in dancing skills is understood as the absolute value of difference of a_i variable. Your task is to find out what pairs and in what order will start dancing.

Input

The first line contains an integer n (1 ≤ n ≤ 2·10⁵) — the number of people. The next line contains n symbols B or G without spaces. Bstands for a boy, G stands for a girl. The third line contains n space-separated integers a_i (1 ≤ a_i ≤ 10⁷) — the dancing skill. People are specified from left to right in the order in which they lined up.

Output

Print the resulting number of couples k. Then print k lines containing two numerals each — the numbers of people forming the couple. The people are numbered with integers from 1 to n from left to right. When a couple leaves to dance you shouldn't renumber the people. The numbers in one couple should be sorted in the increasing order. Print the couples in the order in which they leave to dance.

Sample test(s)

input

4
BGBG
4 2 4 3

output

2
3 4
1 2

input

4
BBGG
4 6 1 5

output

2
2 3
1 4

input

4
BGBB
1 1 2 3

output

1
1 2

思路：使用数组 ll[],rr[] 存当前点的左点和右点,然后先遍历一遍存入所有符合条件的数据

class node
{
  public:int u,v,c;男生女生的位置，以及最小技术差
};

然后将这些都压入优先队列中，从中取出没有遍历过的两点，然后更新那个点，符合条件就压入队列。

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int mm=2e5+9;
class node
{
  public:int u,v,c;
};
inline bool operator <(const node a,const node b)
{
  if(a.c^b.c)return a.c>b.c;
  else return a.u>b.u;
}
int ll[mm],rr[mm],skill[mm];
char s[mm];
int n;
int abs(int x)
{
  if(x<0)x=-x;
  return x;
}
bool vis[mm];
priority_queue<node>q;
int main()
{
  while(cin>>n)
  {
    cin>>s+1;
    node z;int num=0;
    for(int i=1;i<=n;i++)
      ll[i]=i-1,rr[i]=i+1,num+=s[i]=='B';
    num=num<n-num?num:n-num;
    for(int i=1;i<=n;i++)
      cin>>skill[i];
      while(!q.empty())q.pop();
    for(int i=1;i<n;i++)
      if(s[i]!=s[i+1])
      z.u=i,z.v=i+1,z.c=abs(skill[i]-skill[i+1]),q.push(z);
     memset(vis,0,sizeof(vis));
     cout<<num<<"\n";
    while(num--)
    {
      while(!q.empty())
      {
        z=q.top();q.pop();
        if(!vis[z.u]&&!vis[z.v])
          {cout<<z.u<<" "<<z.v<<"\n";
           vis[z.u]=vis[z.v]=1;break;
          }
      }
      int b=ll[z.u],t=rr[z.v];
      rr[b]=t;ll[t]=b;
      z.u=b;z.v=t;z.c=abs(skill[b]-skill[t]);
      if(b>0&&t<=n&&s[b]!=s[t])
        q.push(z);
    }
  }
}