hdu 5023 A Corrupt Mayor's Performance Art 廣州網絡賽 線段樹的區間覆蓋+種類查詢

Problem Description

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.

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.

要查詢的是某個區間的顏色集合，用線段樹維護子區間的顏色集合，用一個int型數據保存顏色的集合，1 << i 位是否爲1表示第i種顏色是否存在。

這場網絡賽的所有題目的輸出都沒有後導空格。

//234MS	10584K
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define lch(x) ((x)<<1)
#define rch(x) ((x)<<1|1)

using namespace std;

const int MAX_N = int(1e6) + 1;
int sum[MAX_N<<2]; //tag[MAX_N<<2];
bool tag[MAX_N<<2];
char str[5];

inline void push_up(int n){
    sum[n] = sum[lch(n)] | sum[rch(n)];
}

inline void push_down(int n){
    if(tag[n]){
        //tag[lch(n)] = tag[rch(n)] = tag[n];
        tag[lch(n)] = tag[rch(n)] = true;
        sum[lch(n)] = sum[rch(n)] = sum[n];
        tag[n] = false;
    }
}

void build(int n, int l, int r){
    tag[n] = false;
    if(l == r){
        //scanf("%d", &sum[n]);
        sum[n] = 1<<2;
        return;
    }
    int mid = (l+r)>>1;
    build(lch(n), l, mid);
    build(rch(n), mid+1, r);
    push_up(n);
}

void modify(int a, int b, int v, int n, int l, int r){
    if(a <= l && r <= b){
        tag[n] = true;
        sum[n] = 1<<v;
        return;
    }
    push_down(n);
    int mid = (l+r)>>1;
    if(mid >= a) modify(a, b, v, lch(n), l, mid);
    if(mid+1 <= b) modify(a, b, v, rch(n), mid+1, r);
    push_up(n);
}

int query(int a, int b, int n, int l, int r){
    if(a <= l && r <= b) return sum[n];
    else{
        push_down(n);
        int mid = (l+r)>>1;
        if(b < mid+1) return query(a, b, lch(n), l, mid);
        else if(mid < a) return query(a, b, rch(n), mid+1, r);
        else return query(a, b, lch(n), l, mid) | query(a, b, rch(n), mid+1, r);
    }
}

int main(){
    int N, M;

    while(~scanf("%d%d", &N, &M)){
        if(N == 0 && M == 0) break;
        build(1, 1, N);

        int a, b, c;
        while(M--){
            scanf("%s", str);
            if(str[0] == 'P'){
                scanf("%d%d%d", &a, &b, &c);
                modify(a, b, c, 1, 1, N);
            }
            else if(str[0] == 'Q'){
                scanf("%d%d", &a, &b);
                int res = query(a, b, 1, 1, N);
                bool fir = false;
                for(int i=1; i<=30; i++){
                    if(res & (1<<i)){
                        if(fir) printf(" ");
                        printf("%d", i);
                        fir = true;
                    }
                }
                printf("\n");
            }
        }
    }
    return 0;
}