Just a Hook （区间更新）

最新推荐文章于 2020-11-25 09:23:18 发布

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#线段树

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本文介绍了一种使用线段树解决区间更新问题的方法，通过具体示例解释了如何计算一系列金属棒（铜、银、金）经过多次区间替换后的总价值。

Problem Description

In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.

Now Pudge wants to do some operations on the hook.

Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:

For each cupreous stick, the value is 1.
For each silver stick, the value is 2.
For each golden stick, the value is 3.

Pudge wants to know the total value of the hook after performing the operations.
You may consider the original hook is made up of cupreous sticks.

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases. For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations. Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.

Output

For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.

Sample Input

Sample Output

Case 1: The total value of the hook is 24.

题目大概：

有一个铁链，有很多钩子，分别有铜银金，价值不同，区间更新一段链子的钩子类型，计算最后的总价值。

思路：

用树状数组或线段树，下面是线段树代码。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn = 111111;
int h, w, n;
int biao[maxn << 2];
int sum[maxn << 2];
void pushup(int rt)
{
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void pushdown(int rt, int m)
{
    if (biao[rt])
    {
        biao[rt << 1] = biao[rt << 1 | 1] = biao[rt];
        sum[rt << 1] = (m - (m >> 1))*biao[rt];
        sum[rt << 1 | 1] = (m >> 1)*biao[rt];
        biao[rt] = 0;
    }
}

void build(int l, int r, int rt)
{
    biao[rt] = 0;
    sum[rt] = 1;
    if (l == r)return;
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    pushup(rt);
}

void update(int L, int R, int c, int l, int r, int rt)
{
    if (L <= l&&r <= R)
    {
        biao[rt] = c;
        sum[rt] = c*(r - l + 1);
        return;
    }
    pushdown(rt, r - l + 1);
    int m = (l + r) >> 1;
    if (L <= m)update(L, R, c, lson);
    if (R>m)update(L, R, c, rson);
    pushup(rt);
}

int main()
{
    int t;
    scanf("%d", &t);
    int j = 1;
    while (j<=t)
    {
        int n;
        scanf("%d",&n);
        build(1, n, 1);
        int m;
        scanf("%d",&m);
        for (int i = 1;i <= m;i++)
        {
            int q, w, e;
            scanf("%d%d%d",&q, &w, &e);
            update(q, w, e, 1, n, 1);
        }
        printf("Case %d: The total value of the hook is %d.\n", j, sum[1]);
        j++;
    }
    return 0;
}