25. Events （e起来编程”（武汉网络赛））线段树（区间修改，区间查询）

来源：互联网发布：常用数据处理软件编辑：程序博客网时间：2024/06/03 07:19

Every year, the ACM/ICPC team will hold many contests, some of them are training while others are school contests.

In the year of 2017, there are $n$ contests to be held, and at the beginning of year, we plans the time of each contest.

However, as things are changing. Some other events might affect the time of contest and our team leader wants to know some interesting things about the time of some events.

Input

The first line contains an integer $n$ . ( $\le 10^5$ )

The second line contains $n$ integers, the $i$ -th one of them is $t_i$ , which is the planning time of contest $i$ at the beginning of this year. ( $0\le t_i \le 10^9$ )

The third line contains an integer $q$ . And then $q$ lines follows. ( $\le 10^5$ )

Then each line contains three integers. $L_i$ , $R_i$ , $T_i$ , means the time of $L_i$ -th contest to $R_i$ -th contest will be changed by $T_i$ . And then you should output the earlist time of $L_i$ -th contest to $R_i$ -th contest in the history after the change of time. ( $L_i \le R_i \le n$ , $|T_i| \le 10^9$ )

Output

Output contains $q$ lines, $i$ -th line is the answer after $i$ -th query.

Examples

Input 1

41 2 3 431 2 22 3 33 4 -10

Output 1

12-6

Note

At the beginning, $t_i$ are $(1, 2, 3, 4)$ . After the first change on time, $t_i$ becomes $(3, 4, 3, 4)$ , and then $(3, 7, 6, 4)$ , and at last becomes $(3, 7, - 4, - 6)$ .

Be careful that as the times are relative times, so they can be negative.

题意：n场比赛，m次操作，每场比赛有个初值，操作每次把一个区间的比赛时间修改，每次修改完需要统计这个区间历史的时间最小值。

分析：线段树问题，区间修改和区间询问。

坑点：区间问题需要一个延迟标记 cur，如果多次对一个区间进行覆盖，延迟标记cur多次累加，而下面的节点没有随之进行更新就会出现无法更新到最小值。

比如这个测试数据：

0 0 0 0 0 0 0 0

1 2 -1

1 4 -1

1 4 1

2 2 0

最后的一组答案是-2

处理：首先是变量，每个节点需要有一个当前最小值m、一个历史最小值 lm，一个延迟标记累加和（和指的是延迟标记累加的得到的结果） cur，一个用来保存当cur多次累加时出现的最小和 mcur，一个延迟标记累加前的m值用变量 sm 储存。当 rt 节点 pushdown时，需要更新 rt 节点左右孩子节点lc和rc。更新lc节点的mcur时，mcur就等于min（cur（lc）+mcur（rt），mcur（lc）），然后 lm（lc） =min（mcur（lc）+sm（lc），lm（lc）），接着更新lc节点的cur和m，cur（lc）+=cur（rt），m（lc）+=cur（rt）；

代码;

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <queue>#define mem(p,k) memset(p,k,sizeof(p));#define rep(a,b,c) for(int a=b;a<c;a++)#define pb push_back#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define inf 0x6fffffff#define ll long longusing namespace std;const ll maxx=99999999999999999;struct SD{    ll sm,m,lm,cur,mcur;}tree[110001<<2];//vector<ll> mcur[110000<<2];//第一次我用数组来保存每一个cur，然后向下更新时分析两个cur数组来更新lm，然后超内存了，后来想到这样是不需要的。 ll minn;int n;ll kmin(ll a,ll b){//取最小值     if(a>b)return b;    return a;}void pushup(int rt){//每次pushup的时候把m的初值存在sm里。     tree[rt].sm=tree[rt].m=kmin(tree[rt<<1|1].m,tree[rt<<1].m);    tree[rt].lm=kmin(tree[rt].lm,tree[rt].m);}void build(int l,int r,int rt){        if(l==r){        scanf("%lld",&tree[rt].m);        tree[rt].sm=tree[rt].lm=tree[rt].m;        tree[rt].cur=0;        tree[rt].mcur=0;        return;    }    tree[rt].cur=0;    tree[rt].mcur=0;    tree[rt].lm=maxx;    int m=(l+r)>>1;    build(lson);    build(rson);    pushup(rt);}void pushdown(int rt){// 更新mcur你可以想像成 B数组拼接在A数组后面，这时新的mcur就等于求新数组的以第一个数为起点的连续序列最小和。         if(tree[rt].cur||tree[rt].mcur){        ll k=tree[rt].mcur;                tree[rt<<1].mcur=kmin(tree[rt<<1].mcur,tree[rt<<1].cur+k);//tree【rt】.cur就是B数组的和，tree【rt<<1】.cur就是A数组的和，                                                              //更新A数组的mcur需要比较 （B数组的mcur+A数组的cur）。         tree[rt<<1].cur+=tree[rt].cur;        tree[rt<<1].lm=kmin(tree[rt<<1].mcur+tree[rt<<1].sm,tree[rt<<1].lm);//        tree[rt<<1].m+=tree[rt].cur;                tree[rt<<1|1].mcur=kmin(tree[rt<<1|1].mcur,tree[rt<<1|1].cur+k);        tree[rt<<1|1].cur+=tree[rt].cur;        tree[rt<<1|1].lm=kmin(tree[rt<<1|1].mcur+tree[rt<<1|1].sm,tree[rt<<1|1].lm);        tree[rt<<1|1].m+=tree[rt].cur;        tree[rt].cur=0;        tree[rt].mcur=0;    }}void update(int L,int R,ll c,int l,int r,int rt){    if(L<=l&&r<=R){        tree[rt].cur+=c;        tree[rt].mcur=kmin(tree[rt].mcur,tree[rt].cur);        tree[rt].m+=c;        tree[rt].lm=kmin(tree[rt].mcur+tree[rt].sm,tree[rt].lm);                return ;    }    pushdown(rt);    int m=(l+r)>>1;    if(L<=m)update(L,R,c,lson);    if(m<R)update(L,R,c,rson);    pushup(rt);    }void query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R){                minn=kmin(minn,tree[rt].lm);        return ;    }    int m=(l+r)>>1;    pushdown(rt);    if(L<=m)  query(L,R,lson);    if(m<R) query(L,R,rson);    pushup(rt);        return ;}int main(){    while(scanf("%d",&n)!=EOF){       build(1,n,1);       int q;       cin>>q;       while(q--){           int x,y;           ll l;           scanf("%d%d%lld",&x,&y,&l);           minn=maxx;           update(x,y,l,1,n,1);           query(x,y,1,n,1);          //for(int i=1;i<=9;i++)          //cout<<i<<"=="<<tree[i].cur<<" "<<tree[i].mcur<<" "<<tree[i].m<<"  "<<tree[i].lm<<endl;           cout<<minn<<endl;       }    }    return 0;}/*50215733813 307386685 520261529 450138662 949456829 671225339 934482534 186212823 355799555 356887611 602672843 508909336 210692892 952637814 893261943 469925811 223390088 962250251 597375877 835756337 723643811 562932720 149692260 757876154 704575774 220671023 684366264 738316954 73848422 861265352 458520896 725076433 697684561 979865317 531515637 450141867 282487688 61794196 565099202 795853790 549426545 25149102 703216304 4227519 575161865 376709205 110647653 205336767 369942388 980972085502 28 -96313635410 37 -49367265212 18 3534101306 10 -23803374520 39 2105237439 30 63934737132 36 -919571317 30 -71700848720 41 -3613665113 30 73482794518 39 38598557417 20 -78177961739 49 -4589075022 29 -7970817079 29 26974896710 10 -55895677034 47 -2598705895 16 -41297596435 42 87783825137 43 7356438427 10 21491531629 34 10773890221 45 -3164676719 37 3572108576 18 -59390510041 45 -38241790915 28 -4979289446 22 35945274711 48 68709767728 42 -5638557273 14 -6899665036 43 -7181682610 12 53833808817 24 -98714024344 50 -93630875216 28 -16248126014 24 -21740413918 23 1700832972 26 -50882788526 32 -47270128011 13 -31490774623 48 -34189013512 43 -237118353 38 -89302850914 43 6545780475 23 5148770619 14 82776806133 38 -4528582664 21 -17197150846 49 387029760*/

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