POJ 3017 Cut the Sequence

Posted 2020-09-28 lytwajue

POJ 3017 Cut the Sequence

/**
POJ 3017 Cut the Sequence
http://poj.org/problem?id=3017
解题思路：单调队列
解题分析：
dp[i]=dp[j]+max(j+1,i) //限制条件sum(j+1,i)<m
这样全部的最优决策都在区间为(j+1,i)的单调队列中
dp[i]=min(dp[q[i-1]]+a[q[i]])
i为队列中的决策点
*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#define maxn 100100
#define ll long long
using namespace std;
ll n,m;
int a[maxn];
int q[maxn],boq[maxn];
ll dp[maxn];
ll sum[maxn]={0};
//boq[rear] = if  q.size()>1 q[rear-1] else q 区间的前一个位置
multiset<ll> mst;  //队列中存在的“位置”的最优解集合
multiset<ll> ::iterator it;
ll solve(){
    mst.clear();
    int bo=0;//队列区间起始位置的上一个位置
    int head=0,rear=0;
    for(int i=1;i<=n;i++){
        while(sum[i]-sum[bo]>m) bo++;
        while(head<rear && q[head]<=bo){
            it=mst.find(dp[boq[head]]+a[q[head]]);
            mst.erase(it);
            head++;
        }
        while(head<rear && a[q[rear-1]]<=a[i]){
            it=mst.find(dp[boq[rear-1]]+a[q[rear-1]]);
            mst.erase(it);
            rear--;
        }
        q[rear]=i;
        if(head<rear) boq[rear]=q[rear-1];
        else boq[rear]=bo;
        mst.insert(dp[boq[rear]]+a[i]);
        rear++;
        if(boq[head]<bo){
            it=mst.find(dp[boq[head]]+a[q[head]]);
            mst.erase(it);
            boq[head]=bo;
            mst.insert(dp[boq[head]]+a[q[head]]);
        }
        it=mst.begin();
        dp[i]=*it;
    }
    return dp[n];
}
int main(){
    while(~scanf("%lld%lld",&n,&m)){
        int flag=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
            if(a[i]>m) flag=1;
        }
        if(flag) printf("-1\n");
        else {
            ll ans=solve();
            printf("%lld\n",ans);
        }
    }
    return 0;
}