Help problem SegmentTree with lazy

Правка en1, от Duarte, 2016-03-02 17:12:10

Hello everyone, I'm trying to solve the problem http://www.spoj.com/problems/HORRIBLE/, I solved using BIT, but I'm learning about Lazy Propagation and I want to solve using this. I did a code but it is getting WA, and I don't know why. Anyone can help me ?


#include <bits/stdc++.h> using namespace std; typedef long long int lli; typedef vector<lli> vl; class SegmentTree { private: vl st, lazy, A; int n; int left(int p) { return p << 1; } int right(int p) { return (p << 1) + 1; } lli rsq(int p, int L, int R, int i, int j) { if(lazy[p] != 0) { st[p] += (R - L + 1) * lazy[p]; //RMQ = st[p] += lazy[ if(L != R) { lazy[left(p)] += lazy[p]; lazy[right(p)] += lazy[p]; } lazy[p] = 0; } if(i > R || j < L || R < L) return 0LL; if(L >= i && R <= j) return st[p]; return rsq(left(p), L, (L + R) / 2, i, j) + rsq(right(p), (L + R) / 2 + 1, R, i, j); } void updateRange(int p, int L, int R, int i, int j, int newValue) { if(lazy[p] != 0) { st[p] += (R - L + 1) * lazy[p]; //RMQ = st[p] += lazy[p]; if(L != R) { lazy[left(p)] += lazy[p]; lazy[right(p)] += lazy[p]; } lazy[p] = 0; } if(L > R || L > j || R < i) return; if(L >= i && R <= j) { st[p] += (R - L + 1) * newValue; //RMQ = st[p] += value; if(L != R) { lazy[left(p)] += newValue; lazy[right(p)] += newValue; } return; } updateRange(left(p), L, (L + R) / 2, i, j, newValue); updateRange(right(p), (L + R) / 2 + 1, R, i, j, newValue); st[p] = st[left(p)] + st[right(p)]; } public: SegmentTree(int _n) { n = _n; st.assign(4 * n, 0); lazy.assign(4 * n, 0); } lli rsq(int i, int j) { return rsq(1, 0, n - 1, i, j); } void updateRange(int i, int j, int newValue) { updateRange(1, 0, n - 1, i, j, newValue); } }; int main() { int T; scanf("%d", &T); while(T--) { int n, c; scanf("%d %d", &n, &c); SegmentTree st(n); int x, p, q, v; while(c--) { scanf("%d %d %d", &x, &p, &q); if(x == 1) printf("%lld\n", st.rsq(p - 1, q - 1)); else { scanf("%d", &v); st.updateRange(p - 1, q - 1, v); } } } return 0; }
Теги segment tree, lazy propagation

История

 
 
 
 
Правки
 
 
  Rev. Язык Кто Когда Δ Комментарий
en1 Английский Duarte 2016-03-02 17:12:10 2334 Initial revision (published)