POJ 3017 单调队列dp
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。
開始没啥思路,研究了半天,感觉单调队列dp很的精妙,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]),当中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; }