ACM学习历程—HDU 5023 A Corrupt Mayor's Performance Art（广州赛区网赛）（线段树）

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

 

这道题是用线段树来做的。网上找了比较好的模板看了一下，然后自己来写了一下。

 

 

代码：

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <set>
 #define maxn 1000005
 using namespace std;
 struct node
 {
     int lt, rt, val, turn;
 }tree[*maxn];
 set < int > ans;
 void build (int lt, int rt, int id)
 {
     tree[id].lt = lt;
     tree[id].rt = rt;
     tree[id].val = ;
     tree[id].turn = ;
     if (lt == rt)
         return;
     int mid = (lt + rt) >> ;
     build (lt, mid, id << );
     build (mid + , rt, id <<  | );
 }
 void updata (int lt, int rt, int id, int col)
 {
     if (lt <= tree[id].lt && rt >= tree[id].rt)
     {
         tree[id].val = tree[id].turn = col;
         return;
     }
     if (tree[id].turn != )
     {
         tree[id<<].turn = tree[id<<].val = tree[id].turn;
         tree[id<<|].turn = tree[id<<|].val = tree[id].turn;
         tree[id].turn = ;
     }
     int mid = (tree[id].lt + tree[id].rt) >> ;
     if (lt <= mid)
         updata (lt, rt, id<<, col);
     if (rt > mid)
         updata (lt, rt, id<<|, col);
     if (tree[id<<].val == tree[id<<|].val)
         tree[id].val = tree[id<<].val;
     else
         tree[id].val = ;
 }
 void query (int lt, int rt, int id)
 {
     if (lt > tree[id].rt || rt < tree[id].lt)
         return;
     if (tree[id].val != )
     {
         ans.insert(tree[id].val);
         return;
     }
     query (lt, rt, id<<);
     query (lt, rt, id<<|);
 }
 int main()
 {
     //freopen ("test.txt", "r", stdin);
     int n, m;
     while (scanf ("%d%d", &n, &m) != EOF && (m+n) != )
     {
         build (, n, );
         for (int times = ; times < m; ++times)
         {
             char ch;
             int xx, yy, col;
             getchar();
             scanf ("%c%d%d", &ch, &xx, &yy);
             if (ch == 'P')
             {
                 scanf ("%d", &col);
                 updata (xx, yy, , col);
             }
             else
             {
                 query (xx, yy, );
                 set < int > :: iterator it;
                 bool flag = ;
                 for (it = ans.begin(); it != ans.end(); it++)
                 {
                     if (flag)
                         printf (" ");
                     printf ("%d", *it);
                     flag = ;
                 }
                 printf ("\n");
                 ans.clear();
             }
         }
     }
     return ;
 }