Codeforces 52c circular rmq (interval cyclic line segment tree, Interval Update, interval summation)

Reprinted please indicate the source: http://blog.csdn.net/u012860063

Link: http://codeforces.com/problemset/problem/52/C

You are given Circular ArrayA0 ,?A1 ,?...,?AN? -? 1. There are two types of operations with it:

• INC(Lf,?RG,?V)-This operation increases each element on the segment [Lf,?RG] (Inclusively)V;
• Rmq(Lf,?RG)-This operation returns minimal value on the segment [Lf,?RG] (Inclusively ).

Assume segments to be circular, so ifN? =? 5 andLf? =? 3 ,?RG? =? 1, it means the index sequence: 3 ,? 4 ,? 0 ,? 1.

Write Program to process given sequence of operations.

Input

The first line contains integerN(1? ≤?N? ≤? (200000). The next line contains initial state of the array:A0 ,?A1 ,?...,?AN? -? 1 (? -? 106? ≤?AI? ≤? 106 ),AIAre integer. The third line contains integerM(0? ≤?M? ≤? 200000 ),M-The number of operartons. NextMLines contain one operation each. If line contains two integerLf,?RG(0? ≤?Lf,?RG? ≤?N? -? 1) It meansRmqOperation, it contains three IntegersLf,?RG,?V(0? ≤?Lf,?RG? ≤?N? -? 1 ;? -? 106? ≤?V? ≤? (106 )-INCOperation.

Output

For eachRmqOperation Write result for it. please, do not use % LLD specificator to read or write 64-bit integers in C ++. it is preffered to use cout (also you may use % i64d ).

Sample test (s) Input

41 2 3 443 03 0 -10 12 1

Output

100

The Code is as follows:

# Include <cstdio> # include <algorithm> using namespace STD; # define lson L, M, RT <1 # define rson m + 1, R, RT <1 | 1 // The difference between lson and rson indicates that the Left and Right sons of the node represent the root of the current subtree ), that is, the current node # define INF 0x3fffffff _ int64 mmin [5000047] ;__ int64 Col [5000047] ;__ int64 min (_ int64 A, _ int64 B) {if (a <B) return a; return B;} void pushup (INT RT) {// sum [RT] = sum [RT <1] + sum [RT <1 | 1]; mmin [RT] = min (mmin [RT <1], mmin [RT <1 | 1]);} void Pushdown (int rt, int m) {If (COL [RT]) {Col [RT <1] + = Col [RT]; Col [RT <1 | 1] + = Col [RT]; mmin [RT <1] + = Col [RT]; mmin [RT <1 | 1] + = Col [RT]; Col [RT] = 0 ;}} void build (int l, int R, int RT) {Col [RT] = 0; If (L = r) {scanf ("% i64d ", & mmin [RT]); return;} int M = (L + r)> 1; build (lson); Build (rson); pushup (RT );} void Update (int l, int R, int V, int L, int R, int RT) {If (L <= L & R <= r) {Col [RT] + = V; mmin [RT] + = V; return;} Pushdown (RT, R-l + 1 ); int M = (L + r)> 1; if (L <= m) Update (L, R, V, lson); If (r> m) Update (L, r, V, rson); pushup (RT) ;:int64 query (int l, int R, int L, int R, int RT) {// query interval [L, if (L <= L & R <= r) in R] // the current node is completely included in the query interval. Return mmin [RT]; Pushdown (RT, r-l + 1); int M = (L + r)> 1 ;__ int64 ret = inf; If (L <= m) // go to the left ret = min (Ret, query (L, R, lson); If (r> m) // go to the right ret = min (Ret, query (L, r, rson); return ret;} int main () {int n, m, a, B, c; char op; scanf ("% d", & N ); build (1, n, 1); scanf ("% d", & M); While (M --) {scanf ("% d % C", &, & B, & OP); A ++; // converts the interval to 1 to Nb ++; if (a <= B) {If (OP = '') // indicates that the Operation {scanf ("% d", & C); If (C = 0) continue; update (A, B, C, 1, n, 1);} elseprintf ("% i64d \ n", query (A, B, 1, n, 1);} else {If (OP = '') {scanf ("% d", & C); If (C = 0) continue; update (A, N, C, 1, n, 1 ); // The Shard range is A to N and 1 to bupdate (1, B, c, 1, n, 1);} elseprintf ("% i64d \ n ", min (query (A, N, 1, n, 1), query (1, B, 1, n, 1) ;}} return 0 ;}