poj2777(线段树)

标签：ch int 线段 read poj2777 query include define

Count Color

POJ - 2777

<iframe frameborder="0" height="2048px" id="frame-description" scrolling="no" src="https://vjudge.csgrandeur.cn/problem/description/1465294586?1681268590000" style="box-sizing: inherit; height: 1157.39px" width="100%"></iframe>

思路：暴力能过，线段树维护这个区间的颜色，如果是混色则置为1，如果是单一颜色则设为这个颜色，修改就是正常的区间修改，区间查询就要变一下。还有题解是用二进制做得，可以学一下。

#define _CRT_SECURE_NO_WARNINGS 1
#include<algorithm>
#include<fstream>
#include<iostream>
#include<cstdio>
#include<deque>
#include<string>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<bitset>
//#include<unordered_map>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 310000
#define N 210100
#define endl '\n'
#define exp 1e-8
#define lc p << 1
#define rc p << 1|1
#define lowbit(x) ((x)&-(x))
const double pi = acos(-1.0);
typedef long long LL;
typedef unsigned long long ULL;
inline LL read() {
	ULL x = 0, f = 1;
	char ch = getchar();
	while (ch < '0' || ch>'9') {
		if (ch == '-')
			f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * f;
}
struct tree
{
	int l, r,v,tag;  //v代表颜色，tag是标记
}tr[4 * N];
int n, t, m;
int ha[50];  
void pushup(int p)
{
	if (tr[lc].v == tr[rc].v)  //如果左右区间颜色一样，则为一样
		tr[p].v = tr[lc].v;
	else tr[p].v = 0; //否则置为0
}
void pushdown(int p)  //标记下传
{
	if (tr[p].tag)
	{
		tr[lc].v = tr[rc].v = tr[p].tag;
		tr[lc].tag = tr[rc].tag = tr[p].tag;
		tr[p].tag = 0;
	}
}
void build(int p, int l, int r)
{
	tr[p].l = l, tr[p].r = r;
	if (l == r) { tr[p].v = 1;  return; }
	int m = l + r >> 1;
	build(lc, l, m);
	build(rc, m + 1, r);
	pushup(p);
}
void update(int p, int x, int y, int k)
{
	if (tr[p].v == k)return;  //一个小小的剪枝，当要修改的区间颜色就是c时，直接返回不用修改
	if (x <= tr[p].l && tr[p].r <= y)
	{
		tr[p].v = k;
		tr[p].tag = k;
		return;
	}
	pushdown(p);
	int m = tr[p].l + tr[p].r >> 1;
	if (x <= m)update(lc, x, y, k);
	if (y > m)update(rc, x, y, k);
	pushup(p);
}
void query(int p, int x, int y)
{
	if (tr[p].v!=0)  //如果不为0，则表示为纯色
	{
		if (x <= tr[p].l && tr[p].r <= y)
		{
			ha[tr[p].v] = 1;
			return;
		}
	}
	pushdown(p);
	int m = tr[p].l + tr[p].r >> 1;   //为0就分裂
	if (x <= m) query(lc, x, y);
	if (y > m)  query(rc, x, y);
}
int main()
{
	n = read(), t = read(), m = read();
	build(1, 1, n);
	for (int i = 1; i <= m; i++)
	{
		char a;
		cin >> a;
		if (a == 'C')
		{
			int x, y, c;
			x = read(), y = read(), c = read();
			if (x > y)swap(x, y);
			update(1, x, y, c);
		}
		else if(a=='P')
		{
			int x, y;
			x = read(), y = read();
			if (x > y)swap(x, y);
		    int ans = 0;
			memset(ha, 0, sizeof(ha));
			query(1, x, y);
			for (int i = 1; i <= 40; i++)
			{
				if (ha[i])ans++;
			}
			printf("%d\n", ans);
		}
	}
	return 0;
}

标签：ch,int,线段,read,poj2777,query,include,define
From： https://www.cnblogs.com/wyh344866/p/17320209.html