12: Challenge 5 (directly modify the line segment tree), challenge Line Segment

12: Challenge 5 (directly modify the line segment tree), challenge Line Segment

Total time limit:

10000 ms

 

Time limit for a single test point:

1000 ms

 

Memory limit:

262144kB

Description

For an N-length series with M operations, each operation is one of the following two types:

(1) Change a continuous segment to a number at the same time

(2) Calculate the sum of a continuous segment of a Series

Input

The first line has two integers, N and M.
The integer of N in the second row represents this series.
In the next M line, each line starts with one character. If the character is 'M', it indicates a modification operation. The next three integers x, y, and z are represented in [x, y] the number of this interval is changed to z. If the character is 'Q', it indicates a query operation. The next two integers x and y represent [x, y] the sum of the range.

Output

Output a single line for each query operation, indicating the answer.

Sample Input

5 31 2 3 4 5Q 1 5M 2 3 2Q 3 5

Sample output

1511

Prompt

1 <= N <= 10 ^ 5, 1 <= M <= 10 ^ 5, the input is valid, and All integers and answers can be stored with 32-bit integer characters.

Directly modify the line segment tree,

We should first consider maintaining a modification tag,

Note that this mark can be overwritten every time!

Then change the value range directly.

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # define ls k <1 5 # define rs k <1 | 1 6 using namespace std; 7 const int MAXN = 100001; 8 const int maxn = 0x7ffff; 9 void read (int & n) 10 {11 char c = '+'; int x = 0; bool flag = 0; 12 while (c <'0' | c> '9') {c = getchar (); if (c = '-') flag = 1;} 13 while (c> = '0' & c <= '9') 14 x = (x <1) + (x <3) + c-48, c = getchar (); 15 flag = 1? N =-x: n = x; 16} 17 int n, m; 18 int ans = 0; 19 struct node 20 {21 int l, r, w, f; 22 node () 23 {24 l = r = w = 0; 25 f =-maxn; 26} 27} tree [MAXN <2]; 28 void update (int k) 29 {30 tree [k]. w = tree [ls]. w + tree [rs]. w; 31} 32 void build (int ll, int rr, int k) 33 {34 tree [k]. l = ll; tree [k]. r = rr; 35 if (ll = rr) 36 {37 read (tree [k]. w); 38 return; 39} 40 int mid = (ll + rr)> 1; 41 build (ll, mid, ls); 42 build (mid + 1, rr, rs ); 43 update (k); 44} 45 void push (int k) 46 {47 tree [ls]. w = (tree [ls]. r-tree [ls]. l + 1) * tree [k]. f; 48 tree [rs]. w = (tree [rs]. r-tree [rs]. l + 1) * tree [k]. f; 49 tree [ls]. f = tree [k]. f; 50 tree [rs]. f = tree [k]. f; 51 tree [k]. f =-maxn; 52 53} 54 void change (int k, int wl, int wr, int v) 55 {56 if (wr <tree [k]. l | wl> tree [k]. r) 57 return; 58 if (wl <= tree [k]. l & tree [k]. r <= wr) 59 {60 tree [k]. w = (tree [k]. r-tree [k]. l + 1) * v; 61 Tree [k]. f = v; 62 return; 63} 64 int mid = (tree [k]. l + tree [k]. r)> 1; 65 if (tree [k]. f! =-Maxn) 66 push (k); 67 change (ls, wl, wr, v); 68 change (rs, wl, wr, v); 69 update (k ); 70} 71 void ask (int k, int wl, int wr) 72 {73 if (wr <tree [k]. l | wl> tree [k]. r) 74 return; 75 if (wl <= tree [k]. l & tree [k]. r <= wr) 76 {77 ans + = tree [k]. w; 78 return; 79} 80 int mid = (tree [k]. l + tree [k]. r)> 1; 81 if (tree [k]. f! =-Maxn) 82 push (k); 83 ask (ls, wl, wr); 84 ask (rs, wl, wr); 85 update (k ); 86} 87 int main () 88 {89 read (n); read (m); 90 build (1, n, 1); 91 for (int I = 1; I <= m; I ++) 92 {93 char c; int x, y; 94 cin> c; 95 read (x); read (y ); 96 if (c = 'M') 97 {98 int v; 99 read (v); 100 change (1, x, y, v ); 101} 102 else103 {104 ans = 0; 105 ask (1, x, y); 106 printf ("% d \ n", ans ); 107} 108} 109 return 0; 110} 111 12: Challenge 5 recent submission