POJ-2777-CountColor(线段树，位运算）

最新推荐文章于 2019-10-17 17:01:06 发布

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最新推荐文章于 2019-10-17 17:01:06 发布

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原文链接：http://www.cnblogs.com/YDDDD/p/10847767.html

本文介绍了一种利用线段树数据结构解决颜色区间查询和更新的问题，通过位运算实现颜色数量的快速计算。文章详细解释了线段树的构建、更新和查询操作，并提供了完整的代码示例。

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链接：https://vjudge.net/problem/POJ-2777#author=0

题意：

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:

1. "C A B C" Color the board from segment A to segment B with color C.
2. "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.

思路：

线段树，还是普通的线段树，染色的查询和更新使用位运算，因为颜色区间在（1-30）之内。

所以可以使用(1<<1-1<<30)来表示这中二进制1的个数来表示颜色的数量。

不过我之前的写的普通的线段树我也不知道为啥会WA。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <memory.h>
#include <algorithm>
#include <string>
#include <stack>
#include <vector>
#include <queue>

using namespace std;
typedef long long LL;
const int MAXN = 1e5+10;

int Seg[MAXN*4];
int lazy[MAXN*4];
int Vis[100];
int n, t, o;
int res;

void PushDown(int root)
{
    if (lazy[root] != 0)
    {
        Seg[root<<1] = (1<<lazy[root]);
        Seg[root<<1|1] = (1<<lazy[root]);

        lazy[root<<1] = lazy[root];
        lazy[root<<1|1] = lazy[root];
        lazy[root] = 0;
    }
}

void PushUp(int root)
{
    Seg[root] = Seg[root<<1]|Seg[root<<1|1];
}

void Build(int root, int l, int r)
{
    if (l == r)
    {
        Seg[root] = 2;
        return;
    }
    int mid = (l + r) / 2;
    Build(root << 1, l, mid);
    Build(root << 1 | 1, mid + 1, r);
    PushUp(root);
}

void Update(int root, int l, int r, int ql, int qr, int c)
{
    if (r < ql || qr < l)
        return;
    if (ql <= l && r <= qr)
    {
        Seg[root] = (1<<c);
        lazy[root] = c;
        return;
    }
    PushDown(root);
    int mid = (l+r)/2;
    Update(root<<1, l, mid, ql, qr, c);
    Update(root<<1|1, mid+1, r, ql, qr, c);
    PushUp(root);
}

int Query(int root, int l, int r, int ql, int qr)
{
    if (r < ql || qr < l)
        return 0;
    if (ql <= l && r <= qr)
    {
        return Seg[root];
    }
    int mid = (l+r)/2;
    PushDown(root);
    int col1 = 0, col2 = 0;
    col1 = Query(root<<1, l, mid, ql, qr);
    col2 = Query(root<<1|1, mid+1, r, ql, qr);
    return col1|col2;
}

int Get(int x)
{
    int res = 0;
    while (x)
    {
        if (x&1)
            res++;
        x >>= 1;
    }
    return res;
}

int main()
{
    char op[10];
    int a, b, c;
    while (~scanf("%d%d%d", &n, &t, &o))
    {
        Build(1, 1, n);
        while (o--)
        {
            scanf("%s", op);
            if (op[0] == 'C')
            {
                scanf("%d%d%d", &a, &b, &c);
                if (a > b)
                    swap(a, b);
                Update(1, 1, n, a, b, c);
            }
            else
            {
                scanf("%d%d", &a, &b);
                if (a > b)
                    swap(a, b);
                memset(Vis, 0, sizeof(Vis));
                int res = Query(1, 1, n, a, b);
                printf("%d\n", Get(res));
            }
        }
    }

    return 0;
}

转载于:https://www.cnblogs.com/YDDDD/p/10847767.html