题目传送：Just a Hook
思路：线段树，成段替换， 区间求和。成段更新时，注意延迟标记的作用，它就是用来暂停往下更新来达到节省时间的，然后每次更新每个节点的子节点之前都要判断是否需要往下更新。
AC代码：
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <deque>#include <cctype>#define LL long long#define INF 0x7fffffffusing namespace std;const int maxn = 100005;int sum[maxn << 2];//求区间和 int lazy[maxn << 2];//延迟标记 void pushdown(int rt, int m) {if(lazy[rt]) {//如果之前这里做了标记，则说明没有往下更新，暂停了一下，，用来判断是否需要往下更新 lazy[rt << 1] = lazy[rt << 1 | 1] = lazy[rt];sum[rt << 1] = (m – (m >> 1)) * lazy[rt];sum[rt << 1 | 1] = (m >> 1) * lazy[rt];lazy[rt] = 0;//往下更新完后，标记置为0，即当前不需要往下更新 }}void build(int l, int r, int rt) {lazy[rt] = 0;sum[rt] = r – l + 1;if(l == r) return;int mid = (l + r) >> 1;build(l, mid, rt << 1);build(mid + 1, r, rt << 1 | 1);}void update(int L, int R, int c, int l, int r, int rt) {if(L <= l && r <= R) {sum[rt] = c * (r – l + 1);lazy[rt] = c;//延迟标记，每次把该段更新完后暂时不往下更新，节省时间 return;}pushdown(rt, r – l + 1);//向下更新 int mid = (l + r) >> 1;if(L <= mid) update(L, R, c, l, mid, rt << 1);if(R >= mid + 1) update(L, R, c, mid + 1, r, rt << 1 | 1);sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];//向上更新 }int main() {int T, n, m;scanf("%d", &T);for(int cas = 1; cas <= T; cas ++) {scanf("%d %d", &n, &m);build(1, n, 1);for(int i = 0; i < m; i ++) {int a, b, c;scanf("%d %d %d", &a, &b, &c);update(a, b, c, 1, n, 1);}printf("Case %d: The total value of the hook is %d.\n", cas, sum[1]);}return 0;}
梦想让我与众不同，奋斗让我改变命运！
