2527: [Poi2011]Meteors
Time Limit: 60 Sec Memory Limit: 128 MB
Submit: 405 Solved: 160
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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
Source
整体二分。
其实就是分治的思想，我们把原问题分成若干个形式相同，规模更小的子问题，这里是两个，所以叫做“二分”。
我们把能在mid次之前收集完的国家分成一组，把前mid次的流星雨传进去，就是原问题的结构。
把后mid次的国家分成一组，把后mid次的流星雨传进去，把每个国家需要的减去其所在前mid次中得到的。
如何快速计算每个国家得到的呢？
线段树和树状数组都太慢了，我们对于这种区间修改，单点查询的操作可以采用差分的思想。
后来发现还是树状数组快
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #define maxn 300010 #define INF 100000000000ll using namespace std; long long c1[maxn],c2[maxn]; int n,m,p; int ans[maxn],to[maxn],Head[maxn],Next[maxn],num=0; struct note{ int pos; long long cur,need; }q[maxn],q1[maxn],q2[maxn]; void make_way(int u,int v) { to[++num]=v; Next[num]=Head[u]; Head[u]=num; } struct rock{ int l,r; long long s; }c[maxn]; #define lowbit(x) x&(-x) //TLETLETLETLRTLR //TLETLETLETLRTLR //TLETLETLETLRTLR //TLETLETLETLRTLR //TLETLETLETLRTLR //TLETLETLETLRTLR //TLETLETLETLRTLR void update(int v,int val) { for(int pp=v;pp<=maxn;c1[pp]+=val,pp+=lowbit(pp)); } long long query(int l,int r) { long long ans=0; for(int pp=r;pp>0;ans+=c1[pp],pp-=lowbit(pp)); return ans; } void solve(int l,int r,int head,int tail) { if(head>tail)return; if(l==r) { for(int i=head;i<=tail;i++) { ans[q[i].pos]=l; } return; } int mid=(l+r)>>1,t1=0,t2=0; for(int i=l;i<=mid;i++) if(c[i].l<=c[i].r) { update(c[i].l,c[i].s); update(c[i].r+1,-c[i].s); }else { update(c[i].l,c[i].s); update(m+1,-c[i].s); update(1,c[i].s); update(c[i].r+1,-c[i].s); } for(int i=head;i<=tail;i++) { long long tmp=0; for(int j=Head[q[i].pos];j&&tmp<INF;j=Next[j]) { tmp+=query(j,j); } if(q[i].cur+tmp>=q[i].need) q1[++t1]=q[i];else { q[i].cur+=tmp; q2[++t2]=q[i]; } } for(int i=l;i<=mid;i++) if(c[i].l<=c[i].r) { update(c[i].l,-c[i].s); update(c[i].r+1,c[i].s); }else { update(c[i].l,-c[i].s); update(m+1,c[i].s); update(1,-c[i].s); update(c[i].r+1,c[i].s); } for(int i=1;i<=t1;i++) q[head+i-1]=q1[i]; for(int i=1;i<=t2;i++) q[head+t1+i-1]=q2[i]; solve(l,mid,head,head+t1-1); solve(mid+1,r,head+t1,tail); } int main() { scanf("%d %d",&n,&m); for(int i=1;i<=m;i++) { int a; scanf("%d",&a); make_way(a,i); } for(int i=1;i<=n;i++) { scanf("%lld",&q[i].need); q[i].pos=i; q[i].cur=0; } scanf("%d",&p); for(int i=1;i<=p;i++) { scanf("%d %d %d",&c[i].l,&c[i].r,&c[i].s); } c[++p].l=1,c[p].r=m,c[p].s=1000000010; solve(1,p,1,n); for(int i=1;i<=n;i++) { if(ans[i]==p)printf("NIE "); else printf("%d ",ans[i]); } return 0; }
T一个点，不会改
在LHQ大佬的帮助下，我知道了mid不一定每次都要加上去，可以用莫队的思想，每次在先前的基础上加减一些流星雨，这么做看上去是快了一点。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #define maxn 300010 #define INF 100000000000ll using namespace std; long long c1[maxn],c2[maxn]; int n,m,p,now=0; int ans[maxn],to[maxn],Head[maxn],Next[maxn],num=0; struct note{ int pos; long long cur,need; }q[maxn],q1[maxn],q2[maxn]; void make_way(int u,int v) { to[++num]=v; Next[num]=Head[u]; Head[u]=num; } struct rock{ int l,r; long long s; }c[maxn]; #define lowbit(x) x&(-x) void update(int v,int val) { for(int pp=v;pp<=maxn;c1[pp]+=val,pp+=lowbit(pp)); } long long query(int l,int r) { long long ans=0; for(int pp=r;pp>0;ans+=c1[pp],pp-=lowbit(pp)); return ans; } void solve(int l,int r,int head,int tail) { if(head>tail)return; if(l==r) { for(int i=head;i<=tail;i++) { ans[q[i].pos]=l; } return; } int mid=(l+r)>>1,t1=0,t2=0; while(now<mid) { now++; if(c[now].l<=c[now].r) { update(c[now].l,c[now].s); update(c[now].r+1,-c[now].s); }else { update(c[now].l,c[now].s); update(m+1,-c[now].s); update(1,c[now].s); update(c[now].r+1,-c[now].s); } } while(now>mid) { if(c[now].l<=c[now].r) { update(c[now].l,-c[now].s); update(c[now].r+1,c[now].s); }else { update(c[now].l,-c[now].s); update(m+1,c[now].s); update(1,-c[now].s); update(c[now].r+1,c[now].s); } now--; } for(int i=head;i<=tail;i++) { long long tmp=0; for(int j=Head[q[i].pos];j&&tmp<INF;j=Next[j]) { tmp+=query(j,j); } if(q[i].cur+tmp>=q[i].need) q1[++t1]=q[i];else { //q[i].cur+=tmp; q2[++t2]=q[i]; } } for(int i=1;i<=t1;i++) q[head+i-1]=q1[i]; for(int i=1;i<=t2;i++) q[head+t1+i-1]=q2[i]; solve(l,mid,head,head+t1-1); solve(mid+1,r,head+t1,tail); } int main() { scanf("%d %d",&n,&m); for(int i=1;i<=m;i++) { int a; scanf("%d",&a); make_way(a,i); } for(int i=1;i<=n;i++) { scanf("%lld",&q[i].need); q[i].pos=i; q[i].cur=0; } scanf("%d",&p); for(int i=1;i<=p;i++) { scanf("%d %d %d",&c[i].l,&c[i].r,&c[i].s); } c[++p].l=1,c[p].r=m,c[p].s=1000000010; solve(1,p,1,n); for(int i=1;i<=n;i++) { if(ans[i]==p)printf("NIE "); else printf("%d ",ans[i]); } return 0; }