bzoj 2527 [Poi2011]Meteors 整体二分+树状数组

Description

Byteotian Interstellar Union (BIU) has recently discovered a new planet in a nearby galaxy. The planet is unsuitable for colonisation due to strange meteor showers, which on the other hand make it an exceptionally interesting object of study.
The member states of BIU have already placed space stations close to the planet’s orbit. The stations’ goal is to take samples of the rocks flying by. The BIU Commission has partitioned the orbit into msectors, numbered from 1to m, where the sectors 1and mare adjacent. In each sector there is a single space station, belonging to one of the nmember states.
Each state has declared a number of meteor samples it intends to gather before the mission ends. Your task is to determine, for each state, when it can stop taking samples, based on the meter shower predictions for the years to come.
Byteotian Interstellar Union有N个成员国。现在它发现了一颗新的星球，这颗星球的轨道被分为M份（第M份和第1份相邻），第i份上有第Ai个国家的太空站。
这个星球经常会下陨石雨。BIU已经预测了接下来K场陨石雨的情况。
BIU的第i个成员国希望能够收集Pi单位的陨石样本。你的任务是判断对于每个国家，它需要在第几次陨石雨之后，才能收集足够的陨石。
输入：
第一行是两个数N,M。
第二行有M个数，第i个数Oi表示第i段轨道上有第Oi个国家的太空站。
第三行有N个数，第i个数Pi表示第i个国家希望收集的陨石数量。
第四行有一个数K，表示BIU预测了接下来的K场陨石雨。
接下来K行，每行有三个数Li,Ri,Ai，表示第K场陨石雨的发生地点在从Li顺时针到Ri的区间中（如果Li<=Ri，就是Li,Li+1,…,Ri，否则就是Ri,Ri+1,…,m-1,m,1,…,Li），向区间中的每个太空站提供Ai单位的陨石样本。
输出：
N行。第i行的数Wi表示第i个国家在第Wi波陨石雨之后能够收集到足够的陨石样本。如果到第K波结束后仍然收集不到，输出NIE。
数据范围：
数据范围：
1<=n,m,k<=3*10^5
1<=Pi<=10^9
1<=Ai<10^9

Input

The first line of the standard input gives two integers, n and m(1<=n,m<=3*10^5) separated by a single space, that denote, respectively, the number of BIU member states and the number of sectors the orbit has been partitioned into.
In the second line there are mintegers Oi(1<=Oi<=n) separated by single spaces, that denote the states owning stations in successive sectors.
In the third line there are nintegers Pi(1<=Pi<=10^9) separated by single spaces, that denote the numbers of meteor samples that the successive states intend to gather.
In the fourth line there is a single integer k(1<=k<=3*10^5) that denotes the number of meteor showers predictions. The following klines specify the (predicted) meteor showers chronologically. The i-th of these lines holds three integers Li, Ri, Ai(separated by single spaces), which denote that a meteor shower is expected in sectors Li,Li+1,…Ri (if Li<=Ri) or sectors Li,Li+1,…,m,1,…Ri (if Li>Ri), which should provide each station in those sectors with Aimeteor samples (1<=Ai<10^9).
In tests worth at least 20% of the points it additionally holds that .
Output

Your program should print nlines on the standard output. The i-th of them should contain a single integer Wi, denoting the number of shower after which the stations belonging to the i-th state are expected to gather at least Pi samples, or the word NIE (Polish for no) if that state is not expected to gather enough samples in the foreseeable future.
Sample Input

3 5

1 3 2 1 3

10 5 7

3

4 2 4

1 3 1

3 5 2

Sample Output

3

NIE

1

HINT

传送门
不怎么会整体二分……
这题的话，对答案进行二分，然后暴力判断每个国家是否达到了要求，
如果对于mid的答案，那么达到要求的国家放到[L,mid]去判断，
没达到的放到[mid+1,R]去判断。
具体的实现方法：
用树状数组差分的思想把所有[1,m]间要增加的值处理好，
然后对于每个国家开一个vector，那就可以O(m)统计每个国家，
如何把国家分开放呢。。以下是hzwer的写法：
可以在递归里面增加两维cntl,cntr，
每次归完之后，满足的在[cntl,x]内，另外的在(x,cntr]内，
那就可以继续递归了。。
还是太菜。

注意了达到要求就不用继续累积了，不然会爆long long。。。。
一开始脑抽了几发光荣WA 2次。。。

#include<bits/stdc++.h>#define ll long longusing namespace std;const int     N=300005;int n,m,K,T;int a[N],b[N],L[N],R[N],ANS[N];ll tr[N],num[N],P[N];bool flag[N];vector<int>O[N];void add(int x,ll val){    while (x<=m) tr[x]+=val,x+=x&-x;}ll query(int x){    ll val=0LL;    while (x) val+=tr[x],x-=x&-x;    return val;}void update(int l,int r,ll val){    if (l<=r) add(l,val),add(r+1,-val);     else add(1,val),add(r+1,-val),add(l,val);}void solve(int cntl,int cntr,int l,int r){    if (cntl>cntr) return;    if (l==r){        for (int i=cntl;i<=cntr;i++) ANS[a[i]]=l;        return;    }    int mid=(l+r)>>1;    while (T<=mid) ++T,update(L[T],R[T],num[T]);    while (T>mid) update(L[T],R[T],-num[T]),T--;    int x=0;    for (int i=cntl;i<=cntr;i++){        ll tmp=0LL;int t1=O[a[i]].size();        for (int j=0;j<t1;j++){            tmp+=query(O[a[i]][j]);            if (tmp>=P[a[i]]) break;        }        if (tmp>=P[a[i]]) flag[a[i]]=1,x++;            else flag[a[i]]=0;    }    int c1=cntl,c2=cntl+x;    for (int i=cntl;i<=cntr;i++)        if (flag[a[i]]) b[c1++]=a[i]; else b[c2++]=a[i];    for (int i=cntl;i<=cntr;i++) a[i]=b[i];    solve(cntl,c1-1,l,mid);    solve(c1,c2-1,mid+1,r);}int main(){    scanf("%d%d",&n,&m);    int x;    for (int i=1;i<=m;i++){        scanf("%d",&x);        O[x].push_back(i);    }    for (int i=1;i<=n;i++) scanf("%lld",&P[i]);    scanf("%d",&K);    for (int i=1;i<=K;i++)        scanf("%d%d%lld",&L[i],&R[i],&num[i]);    for (int i=1;i<=n;i++) a[i]=i;    K++,L[K]=1,R[K]=m,num[K]=1e12;    T=0,solve(1,n,1,K);    for (int i=1;i<=n;i++)        if (ANS[i]==K) puts("NIE");            else printf("%d\n",ANS[i]);    return 0;}