首页 > 代码库 > HDU 5023 A Corrupt Mayor's Performance Art （线段树）

HDU 5023 A Corrupt Mayor's Performance Art （线段树）

2024-07-23 14:10:36   217人阅读

A Corrupt Mayor‘s Performance Art

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 255    Accepted Submission(s): 114


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
 

题意：
在片段上着色，有两种操作，如下：
第一种：P a b c 把 a 片段至 b 片段的颜色都变为 c 。
第二种：Q a b 询问 a 片段至 b 片段有哪些颜色，把这些颜色按从小到大的编号输入，不要有重复
片段上默认的初始颜色为编号2的颜色。
思路：
1、利用线段树进行区间覆盖。
2、在每个节点上附上一个60大小的数组，便于将左右儿子的颜色编号合并到自己节点的数组上（因为颜色编号最多为30）。
3、每次合并左右儿子的颜色编号都需要一次排序和去重，防止记录数超过数组大小。
4、由于0 <N <= 1,000,000过大，可对所有的 a 和 b 做一次 hash ，减小建树的大小，防止爆内存。

/*************************************************************************
  > File Name: hdu5023.cpp
  > Author: Bslin
  > Mail: Baoshenglin1994@gmail.com
  > Created Time: 2014年09月20日 星期六 21时24分19秒
 ************************************************************************/

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int N = 200010;
const int M = 100010;

struct node {
	int l, r;
	int cnt;
	bool flag;
	int num[70];
} tree[N << 2];
int ans[70], tot;
int Hash[N], a[M], b[M], c[M];
char op[M];

void pushup(int p) {
	int i, cnt;
	memset(tree[p].num, 0, sizeof(tree[p].num));
	for(i = 0; i < tree[p << 1].cnt; i ++) {
		tree[p].num[i] = tree[p << 1].num[i];
	}
	int len = tree[p << 1].cnt;
	for(i = 0; i < tree[p << 1 | 1].cnt; i ++) {
		tree[p].num[len + i] = tree[p << 1 | 1].num[i];
	}
	cnt = tree[p << 1].cnt + tree[p << 1 | 1].cnt;
	sort(tree[p].num, tree[p].num + cnt);
	tree[p].cnt = unique(tree[p].num, tree[p].num + cnt) - tree[p].num;
}

void pushdown(int p) {
	if(tree[p].flag) {
		tree[p << 1].flag = tree[p << 1 | 1].flag = 1;
		tree[p].flag = 0;
		memset(tree[p << 1].num, 0, sizeof(tree[p << 1].num));
		memset(tree[p << 1 | 1].num, 0, sizeof(tree[p << 1 | 1].num));
		tree[p << 1].cnt = 1;
		tree[p << 1 | 1].cnt = 1;
		tree[p << 1].num[0] = tree[p].num[0];
		tree[p << 1 | 1].num[0] = tree[p].num[0];
	}
}

void build(int l, int r, int p) {
	tree[p].l = l;
	tree[p].r = r;
	tree[p].flag = 0;
	tree[p].cnt = 1;
	memset(tree[p].num, 0, sizeof(tree[p].num));
	tree[p].num[0] = 2;
	if(l == r) {
		return ;
	}
	int mid = (l + r) >> 1;
	build(l, mid, p << 1);
	build(mid + 1, r, p << 1 | 1);
}

void update(int l, int r, int val, int p) {
	if(tree[p].l == l && tree[p].r == r) {
		memset(tree[p].num, 0, sizeof(tree[p].num));
		tree[p].cnt = 1;
		tree[p].num[0] = val;
		tree[p].flag = 1;
		return ;
	}
	pushdown(p);
	int mid = (tree[p].l + tree[p].r) >> 1;
	if(r <= mid) update(l, r, val, p << 1);
	else if(l > mid) update(l,r,val,p << 1 | 1);
	else {
		update(l, mid, val, p << 1);
		update(mid + 1, r, val, p << 1 | 1);
	}
	pushup(p);
}

void query(int l, int r, int p) {
	if(tree[p].l == l && tree[p].r == r) {
		int i;
		for(i = 0; i < tree[p].cnt; i ++) {
			ans[tot++] = tree[p].num[i];
		}
		sort(ans, ans + tot);
		tot = unique(ans, ans + tot) - ans;
		return ;
	}
	pushdown(p);
	int mid = (tree[p].l + tree[p].r) >> 1;
	if(mid >= r) query(l, r, p << 1);
	else if(l > mid) query(l, r, p << 1 | 1);
	else {
		query(l, mid, p << 1);
		query(mid + 1, r, p << 1 | 1);
	}
}

int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
	freopen("in", "r", stdin);
#endif
	int n, m, i, j;
	int hh;
	while(scanf("%d%d", &n, &m) != EOF) {
		if(n == 0 && m == 0) break;
		hh = 0;
		getchar();
		for(i = 0; i < m; i ++) {
			scanf("%c", &op[i]);
			if(op[i] == 'P') {
				scanf("%d%d%d", &a[i], &b[i], &c[i]);
				Hash[hh++] = a[i];
				Hash[hh++] = b[i];
			} else {
				scanf("%d%d", &a[i], &b[i]);
				Hash[hh++] = a[i];
				Hash[hh++] = b[i];
			}
			getchar();
		}
		sort(Hash, Hash + hh);
		hh = unique(Hash, Hash + hh) - Hash;
		build(1, hh, 1);
		for(i = 0; i < m; i ++) {
			a[i] = lower_bound(Hash, Hash + hh, a[i] ) - Hash + 1;
			b[i] = lower_bound(Hash, Hash + hh, b[i] ) - Hash + 1;
		}
		for(i = 0; i < m; i++) {
			if(op[i] == 'P') {
				update(a[i], b[i], c[i], 1);
			} else {
				memset(ans, 0, sizeof(ans));
				tot = 0;
				query(a[i], b[i], 1);
				for(j = 0; j < tot; j ++) {
					if(j == 0) printf("%d", ans[0]);
					else printf(" %d", ans[j]);
				}
				printf("\n");
			}
		}
	}
}


HDU 5023 A Corrupt Mayor's Performance Art （线段树）


联系
我们