HDU 1698 Just a hook 线段树

简单的线段树区间更新。区间求和。莫名奇妙的是 数组要开3*N才能过。坑了好久。

#include<stdio.h>
#include<string.h>
#include<iostream>
#define maxn 300000 + 10
using namespace std;

int val[maxn];

struct Tree
{
    int mark;
    int total;
    int left, right;
}tree[maxn*4];

int create(int root, int left, int right)  // 以root为根节点建树。
{
    tree[root].mark = 0;
    tree[root].left = left;
    tree[root].right = right;
    if (left == right)
    return tree[root].total = val[left];
    int mid = (left + right) / 2;
    int a, b;
    a = create(2*root, left, mid);
    b = create(2*root+1, mid+1, right);
    return tree[root].total = a + b;
}

void update_mark(int root)
{
    if (tree[root].mark)
    {
        tree[root].total = tree[root].mark * (tree[root].right - tree[root].left + 1);
        if (tree[root].left != tree[root].right)
        tree[root*2].mark = tree[root*2+1].mark = tree[root].mark;
        tree[root].mark = 0;

    }
}

int update(int root, int left, int right, int val)  // 区间更新
{
    update_mark(root);
    if (tree[root].left > right || tree[root].right < left)  // 当前区间和更新区间无交集。不作处理。
    return tree[root].total;
    if (left <= tree[root].left && tree[root].right <= right)  // 没动延迟标记咋用的.这是当前区间完全包含在更新区间的时候。
    {
        tree[root].mark = val;
        return tree[root].total = val * (tree[root].right - tree[root].left + 1);
    }
    int a = update(2*root, left, right, val);
    int b = update(2*root+1, left, right, val);
    return tree[root].total = a + b;
}

int calculate(int root, int left, int right)  // 求区间和
{
    update_mark(root);
    if (tree[root].left > right || tree[root].right < left)
    return 0;
    if (left <= tree[root].left && tree[root].right <= right)
    return tree[root].total;
    int a = calculate(2*root, left, right);
    int b = calculate(2*root+1, left, right);
    return a + b;
}

int main()
{
    int t, n, q;
    scanf("%d", &t);
    int cnt = 0;
    while(t--)
    {
        cnt++;
        scanf("%d%d", &n, &q);
        for (int i=1; i<=n; ++i)
        val[i] = 1;
        create(1, 1, n);
        for (int i=0; i<q; ++i)
        {
            int x, y, te;
            scanf("%d%d%d", &x, &y, &te);
            update(1, x, y, te);
        }
        int ans = calculate(1, 1, n);
        printf("Case %d: The total value of the hook is %d.\n", cnt, ans);
    }
    return 0;
}