CF1157F Maximum Balanced Circle
思路
观察到答案一定是连续的一段下凸函数或者上凸函数
直接模拟找出即可
时间复杂度为\(O(n)\)
代码
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n,a[200200],times[200200],minval=0x3f3f3f3f,maxval=0,ansbegin,anslast,ansnum,beginx,last,num,belong;
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
minval=min(minval,a[i]);
maxval=max(maxval,a[i]);
}
for(int i=1;i<=n;i++)
times[a[i]]++;
for(int i=minval;i<=maxval;i++)
if(times[i]&×[i-1])
ansnum=2,ansbegin=i-1,anslast=i,belong=1;
for(int i=minval-1;i<=maxval+1;i++){
if(times[i]>1){
num+=times[i];
last=i;
if(num>ansnum){
belong=1;
ansnum=num;
ansbegin=beginx;
anslast=last;
}
}
else if(times[i]==1){
num+=times[i];
last=i;
if(num>ansnum){
belong=1;
ansnum=num;
ansbegin=beginx;
anslast=last;
}
num=0;
beginx=i+1;
}
else{
num=0;
beginx=i+1;
}
}//上凸
for(int i=maxval+1;i>=minval-1;i--){
if(times[i]>1){
num+=times[i];
beginx=i;
if(num>ansnum){
belong=2;
ansnum=num;
ansbegin=beginx;
anslast=last;
}
}
else if(times[i]==1){
num+=times[i];
beginx=i;
if(num>ansnum){
belong=2;
ansnum=num;
ansbegin=beginx;
anslast=last;
}
num=0;
last=i-1;
}
else{
num=0;
last=i-1;
}
}// 下凸
if(belong==1&×[ansbegin-1]==1&×[ansbegin]>1)
ansnum++;
if(belong==2&×[anslast+1]==1&×[anslast]>1)
ansnum++;
printf("%d\n",ansnum);
if(belong==1){
if(times[ansbegin-1]==1&×[ansbegin]>1)
printf("%d ",ansbegin-1);
for(int i=ansbegin;i<=anslast;i++){
while(times[i]>1){
printf("%d ",i);
times[i]--;
}
}
for(int i=anslast;i>=ansbegin;i--){
if(times[i])
printf("%d ",i);
}
}
else{
if(times[anslast+1]==1&×[anslast]>1)
printf("%d ",anslast);
for(int i=anslast;i>=ansbegin;i--){
while(times[i]>1){
printf("%d ",i);
times[i]--;
}
}
for(int i=ansbegin;i<=anslast;i++){
if(times[i])
printf("%d ",i);
}
}
return 0;
}