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 0x7fffffff
using 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;
}
HDU - 1698 - Just a Hook (线段树-成段更新)
原文:http://blog.csdn.net/u014355480/article/details/45726057