题目大意:
就是现在一串钩子初始每个部分价值为1, 每次操作修改其中连续的一整段的价值, 问经过Q次操作之后整串钩子的总价值
钩子长度N <= 100000, 操作次数Q <= 100000
大致思路:
就是用懒惰标记来进行延迟更新, 使得每次更新的复杂度保持在O(logn)的级别, 注意一下懒惰标记的下推即可
代码如下:
Result : Accepted Memory : 1684 KB Time : 998 ms
/* * Author: Gatevin * Created Time: 2015/8/16 10:43:12 * File Name: Sakura_Chiyo.cpp */ #include<iostream> #include<sstream> #include<fstream> #include<vector> #include<list> #include<deque> #include<queue> #include<stack> #include<map> #include<set> #include<bitset> #include<algorithm> #include<cstdio> #include<cstdlib> #include<cstring> #include<cctype> #include<cmath> #include<ctime> #include<iomanip> using namespace std; const double eps(1e-8); typedef long long lint; #define maxn 100010 struct Segment_Tree { #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 int val[maxn << 2]; int flag[maxn << 2];//懒惰标记 void pushUp(int rt) { flag[rt] = 0; val[rt] = val[rt << 1] + val[rt << 1 | 1]; return; } void pushDown(int l, int r, int mid, int rt) { if(flag[rt]) { flag[rt << 1] = flag[rt << 1 | 1] = flag[rt]; val[rt << 1] = (mid - l + 1)*flag[rt]; val[rt << 1 | 1] = (r - mid)*flag[rt]; flag[rt] = 0; } return; } void build(int l, int r, int rt) { if(l == r) { val[rt] = 1; flag[rt] = 0; return; } int mid = (l + r) >> 1; build(lson); build(rson); pushUp(rt); return; } void update(int l, int r, int rt, int L, int R, int value) { if(l >= L && r <= R) { val[rt] = (r - l + 1)*value; flag[rt] = value; return; } int mid = (l + r) >> 1; pushDown(l, r, mid, rt);//向下推懒惰标记 if(mid >= L) update(lson, L, R, value); if(mid + 1 <= R) update(rson, L, R, value); pushUp(rt); } int query(int l, int r, int rt, int L, int R) { if(l >= L && r <= R) return val[rt]; int mid = (l + r) >> 1; pushDown(l, r, mid, rt); int ret = 0; if(mid >= L) ret += query(lson, L, R); if(mid + 1 <= R) ret += query(rson, L, R); return ret; } }; Segment_Tree ST; int main() { int n, T; scanf("%d", &T); for(int cas = 1; cas <= T; cas++) { scanf("%d", &n); ST.build(1, n, 1); int Q; scanf("%d", &Q); while(Q--) { int l, r, v; scanf("%d %d %d", &l, &r, &v); ST.update(1, n, 1, l, r, v); } printf("Case %d: The total value of the hook is %d.\n", cas, ST.query(1, n, 1, 1, n)); } return 0; }