poj 3017（单调队列）

想了半天终于是看的题解，网上的题解好像都是出自一人之手，写的都一毛一样，令人窒息。

建了一个递减的队列，具体很难讲的清，又不想复制别人的讲解，就这样吧。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#define ls 2*rt
#define rs 2*rt+1
#define lson ls,L,mid
#define rson rs,mid+1,R
#define ll long long
using namespace std;
typedef pair<int,int> pii;
const ll inf = 0x3f3f3f3f;
/*void dis(int a[], int n){
	printf("总数为%d个\n",n); 
	for(int i = 0; i < n; i++) 	cout<<a[i]<<", ";
	cout<<endl<<"------------------"<<endl;		
}*/

const int mx = 1e5+10;
ll n,m;
ll p[mx],dp[mx],qu[mx],sum,pos,tail,head;
int ok;
int main(){
	//int T=10;
	while(scanf("%lld%lld",&n,&m) != EOF){
		ok = 0;
		for(int i = 1; i <= n; i++){
			scanf("%lld",p+i);
			if(p[i] > m)
				ok = 1;
		}
		if(ok){
			puts("-1");
			continue;
		}
		p[0] = dp[0] = 0;
		head = qu[0] = 0;
		tail = 1;
		pos = sum = 0;
		
		for(int i = 1; i <= n; i++){
			sum += p[i];
			while(pos< i && sum > m){
				sum -= p[pos++];  
			} 
			//sum -= p[qu[head++]];   
			while(tail >head && qu[head] < pos)
				head++;
			
			while(tail >head && p[i] >= p[qu[tail-1]])
				tail--;
			qu[tail++] = i;
			
		
				
			dp[i] = dp[pos-1] + p[qu[head]];
			for(int j = head; j < tail-1; j++)
				dp[i] = min(dp[i],dp[qu[j]]+p[qu[j+1]]);	
		}
		cout<<dp[n]<<endl; 
	}

	
	return 0;
}

本文已参与「新人创作礼」活动，一起开启掘金创作之路