POJ 2777——线段树Lazy的重要性
POJ 2777 Count Color ——线段树Lazy的重要性
原题
链接：http://poj.org/problem?id=2777
Count Color
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 59087 Accepted: 17651
Description
Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem.
There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment with only one color. We can do following two operations on the board:
- "C A B C" Color the board from segment A to segment B with color C.
- "P A B" Output the number of different colors painted between segment A and segment B (including).
In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the beginning, the board was painted in color 1. Now the rest of problem is left to your.
Input
First line of input contains L (1 <= L <= 100000), T (1 <= T <= 30) and O (1 <= O <= 100000). Here O denotes the number of operations. Following O lines, each contains "C A B C" or "P A B" (here A, B, C are integers, and A may be larger than B) as an operation defined previously.
Output
Ouput results of the output operation in order, each line contains a number.
Sample Input
2 2 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2
Sample Output
2
1
Source
POJ Monthly--2006.03.26,dodo
思路
- 题目大意
一开始我们有一段长为L的序列，其上每个数都为1，之后我们要处理o个操作，一种是将一个区间全部修改为另一个数，另一种是给我们一个区间，要求我们算出这个区间内有多少种不同的数，每个数的值小于30
-
思路
- 1.一看区间问题，首先想到了线段树
- 2.发现种类数量不符合区间加法，但是我想到了set去重
- 3.但是又想到这个去重，会让复杂度变成约 \(O(o*L*log_2n)\)， 这对于1e5量级的L与o是绝对不可以的
- 不过又发现每个数最大为30，那便想到了用二进制来表示一个区间内有多少种数——第几位为1表示这个区间内存在几，而最终我们拆分这个二进制数，统计1的个数就能得出答案。区间合并的时候，用 “或|” 就可以完美地将两个区间的信息合并而不用担心去重的问题。
- 还要注意每次区间操作时，先检查左右端点值，保证l <= r以免被坑
- 1.一看区间问题，首先想到了线段树
-
线段树code：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <set>
using namespace std;
long long ff[400005];
int n, t, oo;
long long qu(int o, int l, int r, int x, int y)
{
if (l >= x && r <= y)
{
return ff[o];
}
int mid = (l + r) >> 1;
long long ans1 = 0;
long long ans2 = 0;
if (x <= mid)
{
ans1 = qu(o << 1, l, mid, x, y);
}
if (y > mid)
{
ans2 = qu(o << 1 | 1, mid + 1, r, x, y);
}
return ans2 | ans1;
}
void ch(int o, int l, int r, int x, int y, int v)
{
if (l == r)
{
ff[o] = (1 << (v - 1));
return;
}
int mid = (l + r) >> 1;
if (x <= mid)
{
ch(o << 1, l, mid, x, y, v);
}
if (y > mid)
{
ch(o << 1 | 1, mid + 1, r, x, y, v);
}
ff[o] = ff[o << 1] | ff[o << 1 | 1];
}
int main()
{
scanf("%d%d%d", &n, &t, &oo);
for (int i = 1; i <= 4 * n; ++i)
{
ff[i] = 1;
}
while (oo--)
{
char cc;
scanf("\n%c", &cc);
if (cc == 'C')
{
int ll, rr, tt;
scanf("%d%d%d", &ll, &rr, &tt);
if (ll > rr)
{
int tt = rr;
rr = ll;
ll = tt;
}
ch(1, 1, n, ll, rr, tt);
}
else
{
int ll, rr;
scanf("%d%d", &ll, &rr);
if (ll > rr)
{
int tt = rr;
rr = ll;
ll = tt;
}
long long cco = qu(1, 1, n, ll, rr);
int co = 0;
while (cco)
{
if (cco & 1)
{
++co;
}
cco >>= 1;
}
printf("%d\n", co);
}
}
return 0;
}
这结束了吗？
![[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-FbBZ8HDF-1588656024858)(1.png)]](https://img-blog.csdnimg.cn/20200505132542301.png)
我一开始感到一阵不解，因为计算次数最多是 o*t 约等于 3e6 的量级
好吧缺了些东西，这就是线段树的lazy标记，因为我们没有设置lazy标记，所以导致实际的修改时间复杂度并不是logn而是n，lazy标记的加入也较为简单，新开4*L的数组空间，在修改操作的区间包含当前区间时，进行标记，并且在其他操作之前进行下放即可，这一操作可以将线段树区间操作的复杂度降为logn
AC代码
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <set>
using namespace std;
int ff[400005], ad[400005];
int n, t, oo;
void pd(int o)
{
ad[o << 1] = ad[o];
ad[o << 1 | 1] = ad[o];
ff[o << 1] = ad[o];
ff[o << 1 | 1] = ad[o];
ad[o] = 0;
}
int qu(int o, int l, int r, int x, int y)
{
if (l >= x && r <= y)
{
return ff[o];
}
if (ad[o])
{
pd(o);
}
int mid = (l + r) >> 1;
int ans1 = 0;
int ans2 = 0;
if (x <= mid)
{
ans1 = qu(o << 1, l, mid, x, y);
}
if (y > mid)
{
ans2 = qu(o << 1 | 1, mid + 1, r, x, y);
}
return ans2 | ans1;
}
void ch(int o, int l, int r, int x, int y, int v)
{
if (l >= x && y >= r)
{
ff[o] = (1 << (v - 1));
ad[o] = (1 << (v - 1));
return;
}
if (ad[o])
{
pd(o);
}
int mid = (l + r) >> 1;
if (x <= mid)
{
ch(o << 1, l, mid, x, y, v);
}
if (y > mid)
{
ch(o << 1 | 1, mid + 1, r, x, y, v);
}
ff[o] = ff[o << 1] | ff[o << 1 | 1];
}
int main()
{
scanf("%d%d%d", &n, &t, &oo);
for (int i = 1; i <= 4 * n; ++i)
{
ff[i] = 1;
}
while (oo--)
{
char cc;
scanf("\n%c", &cc);
if (cc == 'C')
{
int ll, rr, tt;
scanf("%d%d%d", &ll, &rr, &tt);
if (ll > rr)
{
int tt = rr;
rr = ll;
ll = tt;
}
ch(1, 1, n, ll, rr, tt);
}
else
{
int ll, rr;
scanf("%d%d", &ll, &rr);
if (ll > rr)
{
int tt = rr;
rr = ll;
ll = tt;
}
int cco = qu(1, 1, n, ll, rr);
int co = 0;
while (cco)
{
if (cco & 1)
{
++co;
}
cco >>= 1;
}
printf("%d\n", co);
}
}
return 0;
}
数据结构在可行的范围内能优化就要优化，不能抱有侥幸心理而增加自己的WA，用线段树处理区间修改类问题时，一定要考虑能否用标记来优化。
