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，每一部分的和不超过m，求每一部分里最大值和的最小值。
开始没啥思路，研究了半天，感觉单调队列dp非常的精妙，先mark一下，后面慢慢理解吧。
代码：
/* ***********************************************
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++];//区间和小于等于m
while(front<rear&&a[i]>=a[que[rear-1]])rear--;//单调严格递减队列
que[rear++]=i;
while(que[front]<p&&front<rear)front++;//把远离的弹出。
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]]);//枚举队列中的元素，求最优解。
}
cout<<dp[n]<<endl;
}
return 0;
}
POJ 3017 单调队列dp,布布扣,bubuko.com