hdu 5023（线段树区间更新+区间查询）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5023

2014 广州网络赛

A Corrupt Mayor's Performance Art

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 83    Accepted Submission(s): 35

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 10P 1 2 3P 2 3 4Q 2 3Q 1 3P 3 5 4P 1 2 7Q 1 3Q 3 4P 5 5 8Q 1 50 0

 

Sample Output

43 44 744 7 8

 

Source

2014 ACM/ICPC Asia Regional Guangzhou Online

 题意：给你一个区间1~n,然后操作和询问次数一共有m次，每次操作都将一个区间涂上一种颜色（这就意味着现在的涂色行为会覆盖之前的涂色行为），每次查询的是一段区间

所有的种颜色，并从小到大输出。

#include <iostream>#include <stdio.h>#include <string.h>#include <string>#include <cstdio>#include <cmath>#include <algorithm>const int N=1e6+100;using namespace std;int cnt[33],ct;struct node   //节点{ int l,r,c;}st[N*4];void build(int v,int l,int r)  //建树{  st[v].l=l;  st[v].r=r;  st[v].c=0;                        if(l==r)return;  int mid=(l+r)/2;  build(2*v,l,mid);  build(2*v+1,mid+1,r);}void insert(int v,int l,int r,int d) //更新{  if(st[v].l==l&&st[v].r==r)   //找到这个区间后，不需要再往下找了  {    st[v].c=d;                        //记录改变的值    return;  }  if(st[v].c>0)  {    st[2*v].c=st[v].c;    st[2*v+1].c=st[v].c;    st[v].c=0;  }  int mid=(st[v].l+st[v].r)/2;  if(l>mid)      insert(2*v+1,l,r,d);  else if(r<=mid)      insert(2*v,l,r,d);  else     {       insert(2*v,l,mid,d);       insert(2*v+1,mid+1,r,d);     }}void getsum(int v,int l,int r)   //这就开始遍历{  if(st[v].l<=l&&r<=st[v].r)  {     if(st[v].c>0)        {         cnt[ct++]=st[v].c;         return ;        }  }  int mid=(st[v].l+st[v].r)/2;  if(r<=mid)getsum(2*v,l,r);  else if(l>mid)getsum(2*v+1,l,r);  else    {      getsum(2*v,l,mid);      getsum(2*v+1,mid+1,r);    }}int main(){ int n,m; while(scanf("%d%d",&n,&m)!=EOF) {   if(n==0&&m==0)break;   memset(cnt,0,sizeof(cnt));   ct=0;   build(1,1,n);   st[1].c=2;   int l,r,d;   char option[10];   while(m--)   {    scanf("%s",option);    if(option[0]=='P')    {      scanf("%d%d%d",&l,&r,&d);      insert(1,l,r,d);    }    else if(option[0]=='Q')    {      scanf("%d%d",&l,&r);      getsum(1,l,r);      sort(cnt,cnt+ct);      for(int i=0;i<ct;i++)      {        if(i==0)printf("%d",cnt[i]);        else        {          if(cnt[i]!=cnt[i-1])          printf(" %d",cnt[i]);        }      }      printf("\n");      memset(cnt,0,sizeof(cnt));      ct=0;    }   } } return 0;}