UESTC360Another LCIS

UESTC360 Another LCIS:http://acm.uestc.edu.cn/#/problem/show/360

Another LCIS

Time Limit: 1000MS Memory Limit: 65535KB 64bit IO Format: %lld & %llu

SubmitStatus

Description

For a sequence        

, and a pair of integers  , if   and            , then the sequence        

is a CIS (Continuous Increasing Subsequence). The longest CIS of a sequence is called theLCIS (Longest Continuous Increasing Subsequence).

In this problem, we will give you a sequence first, and then some add operations and somequery operations. An add operation adds a value to each member in a specified interval. For a query operation, you should output the length of the LCIS of a specified interval.

Input

The first line of the input is an integer  

, which stands for the number of test cases you need to solve.

Every test case begins with two integers  

,  , where   is the size of the sequence, and   is the number of queries.         are specified on the next line, and then   queries follow. Every query begins with a character a or q.a is followed by three integers  , , , meaning that add   to members in the interval   (including  , ), and q is followed by two integers  , , meaning that you should output the length of the LCIS of interval  

.

 

;

 

;

 

;

       

.

Output

For every test case, you should output Case #k: on a single line first, where 

indicates the case number and starts at  

. Then for every q query, output the answer on a single line. See sample for more details.

Sample Input

1
5 6
0 1 2 3 4
q 1 4
a 1 2 -10
a 1 1 -6
a 5 5 -4
q 2 3
q 4 4

Sample Output

Case #1:
4
2
1

Hint

Source

The 9th UESTC Programming Contest Preliminary

思路：虽然都是线段树，但跟上一篇不同……上一篇是单点更新，区间查询LCIS；而这篇难度有所增加，变成区间更新，区间查询LCIS……
显然需要用到lazy标记……但怎么更新呢？当一个区间内的值同时变化时，该区间内的LCIS是不变的，因此这里新开了两个数组lvalue和rvalue,
分别记录区间最左边的值和最右边的值，这样在更新时只需更新这两个值，便可达到相同的效果。其他的跟上一篇的做法大致一样

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <iostream>#include <map>#include <cmath>#include <vector>using namespace std;#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define inf 0x3f3f3f3fconst int maxn= 100000+7;typedef long long ll;int lx[maxn<<2],mx[maxn<<2],rx[maxn<<2],color[maxn<<2];int lvau[maxn<<2],rvau[maxn<<2];int a[maxn];void pushup(int rt,int len) {    lvau[rt] = lvau[rt<<1];    rvau[rt] = rvau[rt<<1|1];    lx[rt] = lx[rt<<1];    rx[rt] = rx[rt<<1|1];    if((lx[rt] == (len-(len>>1))) && (rvau[rt<<1] < lvau[rt<<1|1]))lx[rt] += lx[rt<<1|1];    if((rx[rt] == (len>>1)) && (rvau[rt<<1] < lvau[rt<<1|1]))rx[rt] += rx[rt<<1];    mx[rt] = max(max(mx[rt<<1],mx[rt<<1|1]),(rvau[rt<<1] < lvau[rt<<1|1])?rx[rt<<1]+lx[rt<<1|1]:0);}void build(int l,int r,int rt) {    color[rt] = 0;    lx[rt] = rx[rt] = mx[rt] = 0;    if(l > r)return;    if(l == r) {        scanf("%d",&a[l]);        lvau[rt] = rvau[rt] = a[l];        lx[rt] = mx[rt] = rx[rt] = 1;        return;    }    int mid = l+r>>1;    build(lson);    build(rson);    pushup(rt,r-l+1);}void pushdown(int rt) {    if(color[rt]) {        lvau[rt<<1] += color[rt];        rvau[rt<<1] += color[rt];        lvau[rt<<1|1] += color[rt];        rvau[rt<<1|1] += color[rt];        color[rt<<1] += color[rt];        color[rt<<1|1] += color[rt];        color[rt] = 0;    }}void update(int L,int R,int w,int l,int r,int rt) {    if(L <= l && r <= R) {        color[rt] += w;        lvau[rt] += w;        rvau[rt] += w;        return;    }    if(l > r)return;    pushdown(rt);    int mid = l+r>>1;    if(L <= mid)update(L,R,w,lson);    if(R > mid) update(L,R,w,rson);    pushup(rt,r-l+1);}int query(int L,int R,int l,int r,int rt) {    if(L <= l && r <= R) {        return mx[rt];    }    int mid = l+r>>1;    pushdown(rt);    if(R <= mid)return query(L,R,lson);    else if(L > mid)return query(L,R,rson);    else {        int ltmp = query(L,mid,lson);        int rtmp = query(mid+1,R,rson);        int sum = 0;        if(lvau[rt<<1|1] > rvau[rt<<1]) {            sum = min(mid-L+1,rx[rt<<1])+min(R-mid,lx[rt<<1|1]);        }        return max(sum,max(ltmp,rtmp));    }}int main() {//    freopen("in.txt","r",stdin);//    freopen("out.txt","w",stdout);    int n,m,b,c,t,w,ca = 1;    char ss[10];    scanf("%d",&t);    while(t--) {        scanf("%d%d",&n,&m);        build(1,n,1);        printf("Case #%d:\n",ca++);        for(int i = 1; i <= m; i++) {            scanf("%s",ss);            if(ss[0] == 'a') {                scanf("%d%d%d",&b,&c,&w);                update(b,c,w,1,n,1);            } else {                scanf("%d%d",&b,&c);                printf("%d\n",query(b,c,1,n,1));            }        }    }    return 0;}