POJ 3017 單調佇列dp

_rabbit發表於2014-05-13
原文網址 : https://blog.csdn.net/xianxingwuguan1/article/details/25665641
佇列
Cut the Sequence
Time Limit: 2000MS   Memory Limit: 131072K
Total Submissions: 8764   Accepted: 2576

Description

Given an integer sequence { an } of length N, you are to cut the sequence into several parts every one of which is a consecutive subsequence of the original sequence. Every part must satisfy that the sum of the integers in the part is not greater than a given integer M. You are to find a cutting that minimizes the sum of the maximum integer of each part.

Input

The first line of input contains two integer N (0 < N ≤ 100 000), M. The following line contains N integers describes the integer sequence. Every integer in the sequence is between 0 and 1 000 000 inclusively.

Output

Output one integer which is the minimum sum of the maximum integer of each part. If no such cuttings exist, output −1.

Sample Input

8 17
2 2 2 8 1 8 2 1

Sample Output

12


把序列分成若干部分，每一部分的和不超過m，求每一部分裡最大值和的最小值。

開始沒啥思路，研究了半天，感覺單調佇列dp非常的精妙，先mark一下，後面慢慢理解吧。

程式碼：

/* ***********************************************
Author :_rabbit
Created Time :2014/5/13 1:35:25
File Name :C.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
ll que[100100],a[100100],dp[100100];
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
	 //dp 方程：f[i]=f[j]+max(x[j+1],x[j+2],...,x[i]),其中j<i,x[j+1]+x[j+2]+...+x[i]<=m; 
     ll n,m;
	 while(~scanf("%lld%lld",&n,&m)){
		 bool flag=1;
		 for(int i=1;i<=n;i++){
			 scanf("%lld",&a[i]);
			 if(a[i]>m)flag=0;
		 }
		 if(!flag){
			 puts("-1");continue;
		 }
		 ll front=0,rear=0,p=1;
		 dp[1]=a[1];que[rear++]=1;
		 ll sum=a[1];
		 for(ll i=2;i<=n;i++){
			 sum+=a[i];
			 while(sum>m)sum-=a[p++];//區間和小於等於m
			 while(front<rear&&a[i]>=a[que[rear-1]])rear--;//單調嚴格遞減佇列
			 que[rear++]=i;
			 while(que[front]<p&&front<rear)front++;//把遠離的彈出。
			 dp[i]=dp[p-1]+a[que[front]];
			 for(ll j=front+1;j<rear;j++)
				 dp[i]=min(dp[i],dp[que[j-1]]+a[que[j]]);//列舉佇列中的元素，求最優解。
		 }
		 cout<<dp[n]<<endl;
	 }
     return 0;
}


相關文章