Just a Hook（线段树 + 延迟标记）-CSDN博客

本文链接：https://blog.csdn.net/iteye_6881/article/details/82600604

本文深入探讨了线段树数据结构及其在区间更新问题中的应用，通过实例解析了如何利用线段树高效解决特定类型的问题。重点介绍了线段树的构建、更新和查询方法，以及其在实际编程中的优化策略。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Just a Hook

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16232 Accepted Submission(s): 8075

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

1

10

2

    1 5 2 
  

    5 9 3 
  

Sample Output

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

题意：

给出 T 组 case，后给出 N（1 ~ 100000），代表有 N 个为 1 的数，后给出 M，代表有 M 个操作，每个操作有 l，r，ans，代表将 l 到 r 区间的数变为 ans，在处理完所有操作之后，输出整个序列和的结果。

思路：

线段树 + 区间更新。延迟标记记录每段变为 ans 时候的值，理解了新的写法。

AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX = 100000 * 5;

int tree[MAX], mark[MAX];

void pushup (int node) {
        tree[node] = tree[node << 1] +  tree[node << 1 | 1];
}

void pushdown (int node, int l, int r) {
        if (mark[node]) {
                int mid = (r + l) >> 1;

                mark[node << 1] = mark[node];
                mark[node << 1 | 1] =  mark[node];

                tree[node << 1] = (mid - l + 1) * mark[node];
                tree[node << 1 | 1] = (r - mid) * mark[node];

                mark[node] = 0;
        }
}

void build (int node, int l, int r) {
        int mid = (l + r) >> 1;

        if (l == r) {
                tree[node] = 1;
                mark[node] = 0;
        } else {
                build(node << 1, l, mid);
                build(node << 1 | 1, mid + 1, r);

                pushup(node);
                mark[node] = 0;
        }

}

void update (int node, int l, int r, int cl, int cr, int num) {

        if (cl > r || cr < l) return;
        if (cl <= l && cr >= r) {
                mark[node] = num;
                tree[node] = (r - l + 1) * num;
                return;
        }

        pushdown(node, l, r);
        
        int mid = (r + l) >> 1;
        update(node << 1, l, mid, cl, cr, num);
        update(node << 1 | 1, mid + 1, r, cl, cr, num);

        pushup(node);
}

int main() {
        int t;
        scanf("%d", &t);

        for (int tt = 1; tt <= t; ++tt) {

                int n;
                scanf("%d", &n);
                build(1, 1, n);

                int m;
                scanf("%d", &m);
                while (m--) {
                        int al, ar, ans;
                        scanf("%d%d%d", &al, &ar, &ans);
                        update(1, 1, n, al, ar, ans);
                }

                printf("Case %d: The total value of the hook is %d.\n",
                        tt, tree[1]);

        }

        return 0;
}