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hdu5023A Corrupt Mayor's Performance Art(线段树+位运算）

2024-07-24 00:41:20   220人阅读

题目链接：

huangjing

题意：

有一快板子，然后这个板子被分为从1到n小块，然后给了m个操作，p a b c，是将板子a，b涂成c种颜色，q a b是询问这a到b快板子中有多少种颜色。。
思路：
这个是典型的区间更新，然后涂颜色涉及到位运算，将每一种颜色表示数的各个位，那么这个问题就简单了，，但是我还是一直wa到死，因为我开了一个fbi数组保存结果，结果我把fbi数组开到maxn大小，结果又脑残的对每次询问都进行初始化memset，结果就超时了，还以为自己线段树写挫了。。下次开数组都要小心了。。。
题目：

A Corrupt Mayor‘s Performance Art

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 595    Accepted Submission(s): 239


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
 

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代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=1000000+10;
int tree[maxn<<2],col[maxn<<2];
int fbi[31];
int n,m;

void push_up(int dex)
{
    tree[dex]=tree[dex<<1]|tree[dex<<1|1];
}

void push_down(int dex)
{
    if(col[dex])
    {
        col[dex<<1]=col[dex];
        col[dex<<1|1]=col[dex];
        tree[dex<<1]=col[dex];
        tree[dex<<1|1]=col[dex];
        col[dex]=0;
    }
}

void buildtree(int l,int r,int dex)
{
    col[dex]=0;
    if(l==r)
    {
        tree[dex]=4;
        return;
    }
    int mid=(l+r)/2;
    buildtree(l,mid,dex<<1);
    buildtree(mid+1,r,dex<<1|1);
    push_up(dex);
}

void Update(int L,int R,int dex,int l,int r,int val)
{
    if(l<=L&&r>=R)
    {
        tree[dex]=val;
        col[dex]=val;
        return;
    }
    push_down(dex);
    int mid=(L+R)/2;
    if(l<=mid)  Update(L,mid,dex<<1,l,r,val);
    if(r>mid)  Update(mid+1,R,dex<<1|1,l,r,val);
    push_up(dex);
}

int Query(int L,int R,int dex,int l,int r)
{
    if(l<=L&&r>=R)
       return tree[dex];
    push_down(dex);
    int mid=(L+R)/2;
    if(l>mid)  return Query(mid+1,R,dex<<1|1,l,r);
    else if(r<=mid)  return Query(L,mid,dex<<1,l,r);
    else return Query(L,mid,dex<<1,l,r)|Query(mid+1,R,dex<<1|1,l,r);
}

int main()
{
    char op[2];
    int a,b,c;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)  return 0;
        buildtree(1,n,1);
        for(int j=1;j<=m;j++)
        {
           scanf("%s",op);
           if(op[0]=='P')
           {
               scanf("%d%d%d",&a,&b,&c);
               Update(1,n,1,a,b,1<<c);
           }
           else
           {
               scanf("%d%d",&a,&b);
               int ans=Query(1,n,1,a,b);
               int cnt=0;
               for(int i=1;i<=30;i++)
               {
                    if(ans&(1<<i))
                     fbi[++cnt]=i;
               }
               for(int i=1;i<cnt;i++)
                 printf("%d ",fbi[i]);
               printf("%d\n",fbi[cnt]);
           }
        }
    }
    return 0;
}


hdu5023A Corrupt Mayor's Performance Art(线段树+位运算）


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