hdu----(5023)A Corrupt Mayor's Performance Art(线段树区间更新以及区间查询)

A Corrupt Mayor's Performance Art

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 33    Accepted Submission(s): 11

Problem Description

Corrupt governors always find ways to get dirty money. Paint
something, then sell the worthless painting at a high price to someone
who wants to bribe him/her on an auction, this seemed a safe way for
mayor X to make money.

Because a lot of people praised mayor
X's painting(of course, X was a mayor), mayor X believed more and more
that he was a very talented painter. Soon mayor X was not satisfied with
only making money. He wanted to be a famous painter. So he joined the
local painting associates. Other painters had to elect him as the
chairman of the associates. Then his painting sold at better price.

The local middle school from which mayor X graduated, wanted to beat
mayor X's horse fart(In Chinese English, beating one's horse fart means
flattering one hard). They built a wall, and invited mayor X to paint on
it. Mayor X was very happy. But he really had no idea about what to
paint because he could only paint very abstract paintings which nobody
really understand. Mayor X's secretary suggested that he could make this
thing not only a painting, but also a performance art work.

This was the secretary's idea:

The wall was divided into N segments and the width of each segment
was one cun(cun is a Chinese length unit). All segments were numbered
from 1 to N, from left to right. There were 30 kinds of colors mayor X
could use to paint the wall. They named those colors as color 1, color 2
.... color 30. The wall's original color was color 2. Every time mayor X
would paint some consecutive segments with a certain kind of color, and
he did this for many times. Trying to make his performance art fancy,
mayor X declared that at any moment, if someone asked how many kind of
colors were there on any consecutive segments, he could give the number
immediately without counting.

But mayor X didn't know how to
give the right answer. Your friend, Mr. W was an secret officer of
anti-corruption bureau, he helped mayor X on this problem and gained his
trust. Do you know how Mr. Q did this?

 

Input

There are several test cases.

For each test case:

The first line contains two integers, N and M ,meaning that the wall
is divided into N segments and there are M operations(0 < N <=
1,000,000; 0<M<=100,000)

Then M lines follow, each representing an operation. There are two kinds of operations, as described below:

1) P a b c
a, b and c are integers. This operation means that mayor X painted
all segments from segment a to segment b with color c ( 0 < a<=b
<= N, 0 < c <= 30).

2) Q a b
a and b are
integers. This is a query operation. It means that someone asked that
how many kinds of colors were there from segment a to segment b ( 0 <
a<=b <= N).

Please note that the operations are given in time sequence.

The input ends with M = 0 and N = 0.

 

Output

For each query operation, print all kinds of color on the queried
segments. For color 1, print 1, for color 2, print 2 ... etc. And this
color sequence must be in ascending order.

 

Sample Input

5 10
P 1 2 3
P 2 3 4
Q 2 3
Q 1 3
P 3 5 4
P 1 2 7
Q 1 3
Q 3 4
P 5 5 8
Q 1 5
0 0

 

Sample Output

4
3 4
4 7
4
4 7 8

 

Source

2014 ACM/ICPC Asia Regional Guangzhou Online

 

代码：G++ 281ms

 #define LOCAL
 #include<cstring>
 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #include<iostream>
 using namespace std;
 const int maxn=;

 struct node
 {
     int lef,rig; //通过和的大小来比较种类是否一样
     int cnt; //跟新的种类
     int type; //保存的种类
     int mid(){
       return lef+(rig-lef>>);
     }
 };

 node sac[maxn<<];
 int ans[maxn*],ct;

 void Build(int left,int right,int pos)
 {
   sac[pos]=(node){left,right,,};   //单一
   if(left==right) return ;
   int mid=sac[pos].mid();
   Build(left,mid,pos<<);
   Build(mid+,right,pos<<|);
 }

 void Update(int left,int right,int pos,int val)
 {
    if(left<=sac[pos].lef&&sac[pos].rig<=right)
     {
       sac[pos].cnt=val;
       sac[pos].type=val;
       return ;
     }

     if(sac[pos].cnt!=)
     { //向下更新一次
       sac[pos<<|].cnt=sac[pos<<].cnt=sac[pos].cnt;
       sac[pos<<|].type=sac[pos<<].type=sac[pos].type;
       sac[pos].cnt=;
     }
     int mid=sac[pos].mid();
     if(mid>=left)
        Update(left,right,pos<<,val);
     if(mid<right)
        Update(left,right,pos<<|,val);
       if(sac[pos<<].type==sac[pos<<|].type)
           sac[pos].type=sac[pos<<].type;
         else sac[pos].type=;
 }

 void Query(int pos,int left,int right) {//查找操作

    if(sac[pos].lef>right||sac[pos].rig<left)
         return ;

     if(sac[pos].type)
     {
       ans[ct++]=sac[pos].type;
       return ;
     }
    Query(pos<<,left,right);
    Query(pos<<|,left,right);

 }

 int main()
 {
   int n,m,a,b,c;
   char s[];
   #ifdef LOCAL
      freopen("test.in","r",stdin);
   #endif
   while(scanf("%d%d",&n,&m),n+m!=){
      Build(,n,);
      ct=;
      while(m--)
      {
        scanf("%s",s);
        if(s[]=='P'){
          scanf("%d%d%d",&a,&b,&c);
          Update(a,b,,c);
        }
        else
        {
              scanf("%d%d",&a,&b);
           Query(,a,b);
           sort(ans,ans+ct);
           printf("%d",ans[]);
          for(int i=;i<ct;i++){
             if(ans[i]!=ans[i-])
               printf(" %d",ans[i]);
          }
           printf("\n");
           ct=;
        }
      }
   }
   return ;
 }