pku 3468 A Simple Problem with Integers

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#tree #query #build #hook #struct #2010

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本文详细介绍了解决PKU 3468题目的线段树算法实现过程，包括节点结构定义、构建、更新及查询等核心步骤，并针对乘法溢出问题提出解决方案。

本题明显是线段树的题目，刚开始的时候写的时候，更新操作是使用节点更新，结果果断TLE，然后用树状数组做也果断TLE。后面看了别人的做法，发现要添加一个域用来维护增量，就像hdu1698 Just a Hook需要添加一个域来标记整个区间是否被覆盖了，在查询的时候需要检查增量是否为零，如果不为零则需要把增量往左右子树传递这和hdu1698 Just a Hook的查询有类似的操作，本题有个恶心的地方是中间的乘法会溢出。AC代码如下。

/* Author: ACb0y Date: 2010-9-10 Type: segment tree ProblemId: pku 3468 A Simple Problem with Integers Result: 7601996 ACboy_lrl 3468 Accepted 7196K 1657MS C++ 2470B 2010-09-10 21:54:11 */ #include <iostream> using namespace std; #define MAX 100100 struct node { int left, right; //区间和 long long sum; //增加量 long long add; //求分割点 int get_mid() { return (left + right) / 2; } //求区间长度 int get_dis() { return right - left + 1; } }; node tree[MAX * 3]; long long d[MAX]; //建树 long long build(int left, int right, int loc) { int LL = loc << 1; int RR = LL | 1; tree[loc].left = left; tree[loc].right = right; //初始增量为零 tree[loc].add = 0; if (left == right) { return tree[loc].sum = d[left]; } int mid = tree[loc].get_mid(); return tree[loc].sum = build(left, mid, LL) + build(mid + 1, right, RR); } //更新节点 void update(int left, int right, long long value, int loc) { int LL = loc << 1; int RR = LL | 1; //如果是某个节点，直接增加增量。 if (tree[loc].left == left && tree[loc].right == right) { tree[loc].add += value; return ; } //不管要更新的区间位于什么位置上，根节点的值必定增加((right - left + 1) * value); tree[loc].sum += (long long)(right - left + 1) * value; //递归更新左右节点 int mid = tree[loc].get_mid(); if (right <= mid) { update(left, right, value, LL); } else if (mid < left) { update(left, right, value, RR); } else { update(left, mid, value, LL); update(mid + 1, right, value, RR); } } //查询区间值 long long query(int left, int right, int loc) { int LL = loc << 1; int RR = LL | 1; //刚好查询的区间和是一个节点的区间和 if (tree[loc].left == left && tree[loc].right == right) { return tree[loc].sum + (long long)tree[loc].get_dis() * tree[loc].add; } //区间有增量 if (tree[loc].add) { //先更新本节点的区间和 tree[loc].sum += (long long)tree[loc].get_dis() * tree[loc].add; //增量向左右节点传递 tree[LL].add += tree[loc].add; tree[RR].add += tree[loc].add; //本节点增量设置为零 tree[loc].add = 0; } //递归查询左右子树 int mid = tree[loc].get_mid(); if (right <= mid) { return query(left, right, LL); } else if (mid < left) { return query(left, right, RR); } else { return query(left, mid, LL) + query(mid + 1, right, RR); } } int main() { #ifndef ONLINE_JUDGE freopen("3468.txt", "r", stdin); #endif int n, q; while (cin >> n >> q) { int i; for (i = 1; i <= n; i++) { scanf("%lld", &d[i]); } build(1, n, 1); char op[5]; int a, b, c; while (q--) { scanf("%s", &op); if (op[0] == 'Q') { scanf("%d %d", &a, &b); if (a > b) { int temp = a; a = b; b = temp; } printf("%lld/n", query(a, b, 1)); } else { scanf("%d %d %lld", &a, &b, &c); if (a > b) { int temp = a; a = b; b = temp; } update(a, b, c, 1); } } } return 0; }