算法的理論時(shí)間復(fù)雜度應(yīng)該還是接近 O(n^2)? 但為什么過得還挺快呢 應(yīng)該是后臺數(shù)據(jù)的問題吧
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<vector>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<deque>
#include<algorithm>
#include<cmath>
#define LL long long
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF=0x3f3f3f3f;
const int N=400005;
LL a[N];
int n;
LL m,ans[N];
int dq[N],il,ir;
int main()
{
//freopen("data.in","r",stdin);
scanf("%d %I64d",&n,&m);
for(int i=1;i<=n;++i)
scanf("%I64d",&a[i]);
for(int i=1;i<=n;++i)
if(a[i]>m)
{printf("-1\n");return 0;}
il=0;ir=-1;
LL sum=0;
a[0]=0;
ans[0]=0;
int l=1;
for(int i=1;i<=n;++i)
{
while(il<=ir&&a[dq[ir]]<=a[i])
--ir;
dq[++ir]=i;
sum+=a[i];
while(sum>m)
{
sum-=a[l];
++l;
}
while(dq[il]<l)
{++il;}
if(l==1)
{ans[i]=a[dq[il]];continue;}
ans[i]=ans[l-1]+a[dq[il]];
for(int j=il;j<ir;++j)
{
int k=dq[j];
ans[i]=min(ans[i],ans[k]+a[dq[j+1]]);
}
}
printf("%I64d\n",ans[n]);
return 0;
}