Cut the Sequence
题目
解析
单调队列优化DP好题
万题先拿部分分,我们先考虑裸DP,得到动态转移方程:
dp[i]=min(dp[j]+max(a[k])(j<k<=i))(1<=j<=i)
显然这个O(N2)的DP必T
考虑单调队列优化最大值,AC
1,前缀和优化,O(1)区间求和
2,开longlong
3,有值>m时输出-1
其实听说单调队列优化可以被卡成O(n2),但是A了就行
code:
#include<cstdio>
#include<deque>
#define ll long long
using namespace std;
ll min(ll x,ll y){
return (x<y)?x:y;}
bool idigit(char x){
return (x<'0'|x>'9')?0:1;}
inline ll read()
{
ll num=0,f=1;
char c=0;
while(!idigit(c=getchar())){
if(c=='-')f=-1;}
while(idigit(c))num=(num<<1)+(num<<3)+(c&15),c=getchar();
return num*f;
}
inline void write(ll x)
{
ll F[20];
ll tmp=x>0?x:-x;
if(x<0)putchar('-');
ll cnt=0;
while(tmp>0){
F[cnt++]=tmp%10+'0';tmp/=10;}
while(cnt>0)putchar(F[--cnt]);
if(x==0)putchar('0');
}//快读快输日常
ll n,m,a[100010],s[100010],dp[100010],k,l;
deque <ll> b;
int main()
{
n=read(),m=read();
for(ll i=1;i<=n;i++)
{
a[i]=read(),s[i]=s[i-1]+a[i];//s
if(a[i]>m)
{
printf("-1");
return 0;
}//特判
}
dp[1]=a[1];
k=1;
for(ll i=1;i<=n;i++)
{
while(!b.empty()&&a[i]>=a[b.back()])b.pop_back();
while(s[i]-s[k-1]>m&&k<i)++k;
b.push_back(i);
while(!b.empty()&&b.front()<k)b.pop_front();
dp[i]=dp[k-1]+a[b.front()];
for(int j=1;j<=b.size();j++)
{
l=b.front();
b.pop_front();
if(!b.empty())dp[i]=min(dp[i],dp[l]+a[b.front()]);
b.push_back(l);
}//单调队列优化
}
write(dp[n]);
return 0;
}