2019.8.3 [HZOI] NOIP analog test packet 12 C.
Full match problem solution: https://pan.baidu.com/s/1eSAMuXk
just find it hard to look at this problem,So the partition of the data points
k 1 and 2 only two types are discussed:
- k=1：
According to human intuitionIt is not difficult to think of one kind of greedy strategy: front to back sweep, sweep if the number can be added to the current period to join, or to open a new section.
So you WA ... the
subject of the request lexicographically minimal, and our strategy will let the current period as long as possible , so the division point will be on the list. For example: 12345, can be classified into {1}, {2,3,4}, {5} three sections, and the strategy we obtained the answer is {1,2}, {3,4}, { 5}.
Similar to the "food production" that question, since it does not forward greedy, greedy consider the reverse.
From the forward scan can, based upon the greedy strategy may be as long as possible to ensure that the latter section, i.e. the left as much as possible the division point. Official explanations prove the correctness of this strategy.
Now the question is how can quickly determine whether a number is added to the current paragraph, i.e. a number for B, the current segment is determined whether there is a number a, satisfying \ (A + B = X ^ 2 \) . Found \ (\ sqrt {2 * \ max (a_i)} \) only about 520 may enumerate each x, is determined \ (x ^ 2-b \ ) exists.
Complexity \ (O (n \ sqrt n ) \) - k = 2: still policy described above from the forward sweep, each corresponding to determined whether divided into two sets, so that any two numbers of each set, and after the addition of a number of the current segment is not a perfect square. Similar to the "magic ball" that question is a bipartite graph determination. The bipartite graph is determined by staining high complexity, you can sideband right / disjoint-set extension field determines bipartite graph (similar to the "criminal detention").
The practice k = 1, we think of Merge-value, but then there will be a special case: the time when the sequence is 2,7, 2, and 7 may be divided in two sets; but there are two sequences 2, the two need to be divided in 2 different sets, again a 7 can not be divided in two in the collection. So when the need for special scanning sentence.
Here disjoint-set only write path compression, no merger by rank, so complexity is more than a log.
Code：
#include <bits/stdc++.h>
using namespace std;
const int N=131075;
int table[600]={0,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484,529,576,625,676,729,784,841,900,961,1024,1089,1156,1225,1296,1369,1444,1521,1600,1681,1764,1849,1936,2025,2116,2209,2304,2401,2500,2601,2704,2809,2916,3025,3136,3249,3364,3481,3600,3721,3844,3969,4096,4225,4356,4489,4624,4761,4900,5041,5184,5329,5476,5625,5776,5929,6084,6241,6400,6561,6724,6889,7056,7225,7396,7569,7744,7921,8100,8281,8464,8649,8836,9025,9216,9409,9604,9801,10000,10201,10404,10609,10816,11025,11236,11449,11664,11881,12100,12321,12544,12769,12996,13225,13456,13689,13924,14161,14400,14641,14884,15129,15376,15625,15876,16129,16384,16641,16900,17161,17424,17689,17956,18225,18496,18769,19044,19321,19600,19881,20164,20449,20736,21025,21316,21609,21904,22201,22500,22801,23104,23409,23716,24025,24336,24649,24964,25281,25600,25921,26244,26569,26896,27225,27556,27889,28224,28561,28900,29241,29584,29929,30276,30625,30976,31329,31684,32041,32400,32761,33124,33489,33856,34225,34596,34969,35344,35721,36100,36481,36864,37249,37636,38025,38416,38809,39204,39601,40000,40401,40804,41209,41616,42025,42436,42849,43264,43681,44100,44521,44944,45369,45796,46225,46656,47089,47524,47961,48400,48841,49284,49729,50176,50625,51076,51529,51984,52441,52900,53361,53824,54289,54756,55225,55696,56169,56644,57121,57600,58081,58564,59049,59536,60025,60516,61009,61504,62001,62500,63001,63504,64009,64516,65025,65536,66049,66564,67081,67600,68121,68644,69169,69696,70225,70756,71289,71824,72361,72900,73441,73984,74529,75076,75625,76176,76729,77284,77841,78400,78961,79524,80089,80656,81225,81796,82369,82944,83521,84100,84681,85264,85849,86436,87025,87616,88209,88804,89401,90000,90601,91204,91809,92416,93025,93636,94249,94864,95481,96100,96721,97344,97969,98596,99225,99856,100489,101124,101761,102400,103041,103684,104329,104976,105625,106276,106929,107584,108241,108900,109561,110224,110889,111556,112225,112896,113569,114244,114921,115600,116281,116964,117649,118336,119025,119716,120409,121104,121801,122500,123201,123904,124609,125316,126025,126736,127449,128164,128881,129600,130321,131044,131769,132496,133225,133956,134689,135424,136161,136900,137641,138384,139129,139876,140625,141376,142129,142884,143641,144400,145161,145924,146689,147456,148225,148996,149769,150544,151321,152100,152881,153664,154449,155236,156025,156816,157609,158404,159201,160000,160801,161604,162409,163216,164025,164836,165649,166464,167281,168100,168921,169744,170569,171396,172225,173056,173889,174724,175561,176400,177241,178084,178929,179776,180625,181476,182329,183184,184041,184900,185761,186624,187489,188356,189225,190096,190969,191844,192721,193600,194481,195364,196249,197136,198025,198916,199809,200704,201601,202500,203401,204304,205209,206116,207025,207936,208849,209764,210681,211600,212521,213444,214369,215296,216225,217156,218089,219024,219961,220900,221841,222784,223729,224676,225625,226576,227529,228484,229441,230400,231361,232324,233289,234256,235225,236196,237169,238144,239121,240100,241081,242064,243049,244036,245025,246016,247009,248004,249001,250000,251001,252004,253009,254016,255025,256036,257049,258064,259081,260100,261121,262144,263169,264196,265225,266256,267289,268324,269361,270400};
int n,k,mx,tot,ans[N],a[N];
bool vis[N],dvis[N],issqu[3*N];
void In(int &num){
register char c=getchar();
for(num=0;!isdigit(c);c=getchar());
for(;isdigit(c);num=num*10+c-48,c=getchar());
}
namespace solve1{
inline bool check(int num){//判断num能否加入当前段
for(int i=1;i<=520;++i){
if(table[i]<=num) continue;
if(table[i]-num>mx) break;
if(vis[table[i]-num]) return true;
}
return false;
}
void solve(){
for(int i=n,last=n;i>=2;--i){
vis[a[i]]=true;
mx=max(mx,a[i]);
if(check(a[i-1])){//a[i-1]不能加入当前段
mx=0;
for(int j=last;j>=i;--j) vis[a[j]]=false;//清空数组
ans[++tot]=i-1;//新开一段
last=i-1;
}
}
}
}
namespace solve2{
int fa[3*N];
int Find(int x){
return fa[x]==x?x:fa[x]=Find(fa[x]);
}
bool check(int u,int v){//u和v已经冲突 判断它们能否分到不同集合
int t1=Find(u),t2=Find(u+N);
int s1=Find(v),s2=Find(v+N);
if(t1==s1||t2==s2) return false;
fa[t1]=s2;fa[t2]=s1;
return true;
}
void solve(){
for(int i=1;i*i<=2*N;++i) issqu[i*i]=true;
for(int i=1;i<=2*N;++i) fa[i]=i;
for(int i=n,last=n;i;--i){
if(vis[a[i]]){//出现过
if(issqu[2*a[i]]){
if(dvis[a[i]]) goto fail;//已经出现过两次
for(int j=1;j<=520;++j){
if(table[j]<=a[i]) continue;
if(table[j]-a[i]>mx) break;
if(vis[table[j]-a[i]]&&table[j]!=2*a[i]) goto fail;
}
dvis[a[i]]=true;//标记第二次出现
}
}
else{//没出现过
for(int j=1;j<=520;++j){
if(table[j]<=a[i]) continue;
if(table[j]-a[i]>mx) break;
if(vis[table[j]-a[i]]&&(dvis[table[j]-a[i]]||(!check(a[i],table[j]-a[i])))) goto fail;
}
vis[a[i]]=true;
}
mx=max(mx,a[i]);
continue;
fail:
mx=a[i];
ans[++tot]=i;
for(int j=last;j>=i+1;--j) fa[a[j]]=a[j],fa[a[j]+N]=a[j]+N,vis[a[j]]=dvis[a[j]]=false;
fa[a[i]]=a[i],fa[a[i]+N]=a[i]+N;vis[a[i]]=true;//注意这里 当前段的右端点是i 需要重新标记
last=i;
}
}
}
int main(){
In(n);In(k);
for(int i=1;i<=n;++i) In(a[i]);
if(k==1) solve1::solve();
else solve2::solve();
printf("%d\n",tot+1);
for(int i=tot;i;--i) printf("%d ",ans[i]);
putchar('\n');
return 0;
}