dp（电梯与楼梯）

http://codeforces.com/problemset/problem/1249/E

E. By Elevator or Stairs?

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are planning to buy an apartment in a n

-floor building. The floors are numbered from 1 to n

from the bottom to the top. At first for each floor you want to know the minimum total time to reach it from the first (the bottom) floor.

Let:

• ai

for all i from 1 to n−1 be the time required to go from the i-th floor to the (i+1)-th one (and from the (i+1)-th to the i

• -th as well) using the stairs;
• bi

for all i from 1 to n−1 be the time required to go from the i-th floor to the (i+1)-th one (and from the (i+1)-th to the i-th as well) using the elevator, also there is a value c

• — time overhead for elevator usage (you need to wait for it, the elevator doors are too slow!).

In one move, you can go from the floor you are staying at x

to any floor y (x≠y

) in two different ways:

• If you are using the stairs, just sum up the corresponding values of ai

. Formally, it will take ∑i=min(x,y)max(x,y)−1ai

• time units.
• If you are using the elevator, just sum up c

and the corresponding values of bi. Formally, it will take c+∑i=min(x,y)max(x,y)−1bi

• time units.

You can perform as many moves as you want (possibly zero).

So your task is for each i

to determine the minimum total time it takes to reach the i-th floor from the 1

-st (bottom) floor.

Input

The first line of the input contains two integers n

and c (2≤n≤2⋅105,1≤c≤1000

) — the number of floors in the building and the time overhead for the elevator rides.

The second line of the input contains n−1

integers a1,a2,…,an−1 (1≤ai≤1000), where ai is the time required to go from the i-th floor to the (i+1)-th one (and from the (i+1)-th to the i

-th as well) using the stairs.

The third line of the input contains n−1

integers b1,b2,…,bn−1 (1≤bi≤1000), where bi is the time required to go from the i-th floor to the (i+1)-th one (and from the (i+1)-th to the i

-th as well) using the elevator.

Output

Print n

integers t1,t2,…,tn, where ti is the minimum total time to reach the i

-th floor from the first floor if you can perform as many moves as you want.

Examples

Input

Copy

10 2
7 6 18 6 16 18 1 17 17
6 9 3 10 9 1 10 1 5

Output

Copy

0 7 13 18 24 35 36 37 40 45

Input

Copy

10 1
3 2 3 1 3 3 1 4 1
1 2 3 4 4 1 2 1 3

Output

Copy

0 2 4 7 8 11 13 14 16 17

题意：有n楼，给你1到2，2到3....n-1到n楼的爬楼梯时间和坐电梯时间。从楼梯去坐电梯需要等电梯开门的时间c。问每一楼到一楼的最短时间。

解法：考虑四个转移状态：从楼梯到楼梯，从楼梯到电梯，从电梯到楼梯，从电梯到电梯。

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <stack>;
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 998244353
#define PI acos(-1)
using namespace std;
typedef long long ll ;
ll a[] , b[];
ll dp[][];

int main()
{
    ll n , c ;
    while(~scanf("%lld%lld" , &n , &c))
    {
        for(int i =  ; i <= n ; i++)
        {
            scanf("%lld" , &a[i]);
        }
        for(int i =  ; i <= n ; i++)
        {
            scanf("%lld" , &b[i]);
        }
        memset(dp , INF , sizeof(dp));
        dp[][] = c ;
        dp[][] =  ;
        for(int i =  ; i <= n ; i++)
        {
            dp[i][] = min(dp[i][] , dp[i-][]+a[i]);//电梯到楼梯
            dp[i][] = min(dp[i][] , dp[i-][]+a[i]);//楼梯到楼梯
            dp[i][] = min(dp[i][] , dp[i-][]+b[i]);//电梯到电梯
            dp[i][] = min(dp[i][] , dp[i-][]+b[i]+c);//楼梯到电梯
        }

        for(int i =  ; i < n ; i++)
            cout << min(dp[i][] , dp[i][]) <<  " " ;
        cout << min(dp[n][] , dp[n][]) << endl ;
    }

    return  ;
}