Fzu1608 huge mission line segment tree lazy Interval Update + summation

One of the reasons why I woke up today was that the question was really a dog, it's better to look at the case and guess it quickly,

Question: given some intervals, X, Y. The third parameter W is efficiency, which indicates the unit time efficiency during this period. The total efficiency is (Y-x) * w, then some time periods will be repeated. For example, the first time is 1, 4, 1, and the later time is 2, 4, 3. For a period of, the maximum efficiency must be 3 for the range of 2 and 4, which means that when there is a better situation, we will overwrite the previous one.

Of course, there is another reason for this question: When I used to learn the line segment tree, I used to look at the version provided by the senior student. I didn't look at other people's versions at all, as a result, when I did not understand the meaning of the question last night, I found some details wrong in my mind, because others' versions are different from mine, I did one thing last night. This morning, I handled the small details of my mistakes. I have a long head. Really!

# Include <iostream> # include <cstdio> # include <list> # include <algorithm> # include <cstring> # include <string> # include <queue> # include <stack> # include <map> # include <vector> # include <cmath> # include <memory. h> # include <set> # include <cctype> # define ll long # define ll _ int64 # define EPS 1e-8 # define INF 0 xfffffff // const ll INF = 1ll <<61; using namespace STD; // vector <pair <int, int> G; // typedef pair <int, int> P; // vector <pair <int, int>:: iterator ITER; // Map <LL, int> MP; // Map <LL, int>: iterator P; const int n = 50000 + 5; typedef struct node {int L, R; int num ;}; node tree [N * 4]; void build (int l, int R, int ID) {tree [ID]. L = L; tree [ID]. R = r; tree [ID]. num = 0; If (L = r) return; int mid = (L + r)/2; build (L, mid, id <1 ); build (Mid + 1, R, id <1 | 1);} void Pushdown (int id) {If (tree [ID]. num> 0) {tree [ID <1]. num = max (tree [ID]. num, tree [ID <1]. num); tree [ID <1 | 1]. num = max (tree [ID]. num, tree [ID <1 | 1]. num); tree [ID]. num = 0 ;}} void Update (int l, int R, int L, int R, int ID, int Val) {If (tree [ID]. num> = Val) return; // pruning, if (L <= L & R> = r) {If (tree [ID]. num <Val) tree [ID]. num = val; return;} Pushdown (ID); int mid = (L + r)/2; If (L <= mid) Update (L, R, L, mid, ID <1, Val); If (r> mid) Update (L, R, Mid + 1, R, id <1 | 1, Val ); // else {// Update (L, mid, id <1, Val); // Update (Mid + 1, R, id <1 | 1, Val ); //} int query (int l, int R, int ID) {If (L = r) return tree [ID]. num; Pushdown (ID); int mid = (L + r)/2; int ans1 = 0, ans2 = 0; If (L <= mid) ans1 = query (L, mid, id <1); If (r> mid) ans2 = query (Mid + 1, R, id <1 | 1); Return ans1 + ans2 ;} int main () {int n, m; while (scanf ("% d", & N, & M) = 2) {build (1, n, 1); int q = m; int X, Y, W; while (Q --) {scanf ("% d", & X, & Y, & W); Update (x + 1, Y, 1, n, 1, W);} int ans = query (1, n, 1 ); printf ("% d \ n", ANS);} return 0 ;}

# Include <iostream> # include <cstdio> # include <list> # include <algorithm> # include <cstring> # include <string> # include <queue> # include <stack> # include <map> # include <vector> # include <cmath> # include <memory. h> # include <set> # include <cctype> # define ll long # define ll _ int64 # define EPS 1e-8 # define INF 0 xfffffff // const ll INF = 1ll <<61; using namespace STD; // vector <pair <int, int> G; // typedef pair <int, int> P; // vector <pair <int, int>:: iterator ITER; // Map <LL, int> MP; // Map <LL, int>: iterator P; const int n = 50000 + 5; typedef struct node {int L, R; int num ;}; node tree [N * 4]; int ans; void build (int l, int R, int ID) {tree [ID]. L = L; tree [ID]. R = r; tree [ID]. num = 0; If (L = r) return; int mid = (L + r)/2; build (L, mid, id <1 ); build (Mid + 1, R, id <1 | 1);} void Pushdown (int id) {If (tree [ID]. num> 0) {tree [ID <1]. num = max (tree [ID]. num, tree [ID <1]. num); tree [ID <1 | 1]. num = max (tree [ID]. num, tree [ID <1 | 1]. num); tree [ID]. num = 0 ;}} void Update (int l, int R, int ID, int Val) {If (tree [ID]. num> = Val) return; // pruning, if (L <= tree [ID]. L & R> = tree [ID]. r) {If (tree [ID]. num <Val) tree [ID]. num = val; return;}/* Pushdown (ID); */INT mid = (tree [ID]. L + tree [ID]. r)/2; If (r <= mid) Update (L, R, id <1, Val); else if (L> mid) Update (L, R, ID <1 | 1, Val); else {Update (L, mid, id <1, Val); Update (Mid + 1, R, ID <1 | 1, Val) ;}} void query (int l, int R, int ID) {If (L = r) {ans + = tree [ID]. num; return;} Pushdown (ID); int mid = (tree [ID]. L + tree [ID]. r)/2; If (r <= mid) query (L, R, id <1); else if (L> mid) query (L, R, ID <1 | 1); else {query (L, mid, id <1); query (Mid + 1, R, id <1 | 1 );}} int main () {int n, m; while (scanf ("% d", & N, & M) = 2) {build (1, n, 1); int q = m; int X, Y, W; while (Q --) {scanf ("% d", & X, & Y, & W); Update (x + 1, Y, 1, W);} ans = 0; query (1, n, 1); printf ("% d \ n ", ans);} return 0 ;}