【CodeForces 1249E --- By Elevator or Stairs? 】

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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.

Sample Input

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

Sample Output

0 7 13 18 24 35 36 37 40 45

AC代码：

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
const int MAXN = 200005;
const int INF = 1<<30;
int a[MAXN],b[MAXN];

int main()
{
    SIS;
    int n,c;
    cin >> n >> c;
    for(int i=0;i<n-1;i++) cin >> a[i];
    for(int i=0;i<n-1;i++) cin >> b[i];
    vector<vector<int> > dp(n, vector<int>(2, INF));
    dp[0][0]=0;
    dp[0][1]=c;
    for(int i=0;i<n-1;i++)
    {
        dp[i+1][0]=min(dp[i+1][0],dp[i][1]+a[i]);//在电梯，要走楼梯
        dp[i+1][0]=min(dp[i+1][0],dp[i][0]+a[i]);//不在电梯，要走楼梯
        dp[i+1][1]=min(dp[i+1][1],dp[i][1]+b[i]);//在电梯，要走电梯
        dp[i+1][1]=min(dp[i+1][1],dp[i][0]+b[i]+c);//不在电梯，要走电梯
    }
    for(int i=0;i<n;i++)
        cout << min(dp[i][0],dp[i][1]) << (i==n-1?'\n':' ');
    return 0;
}

#include <iostream>
#include <algorithm>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(),cout.tie()
const int MAXN = 200005;
int a[MAXN],b[MAXN],ans[MAXN];

int main()
{
    SIS;
    int n,c;
    cin >> n >> c;
    for(int i=1;i<n;i++) cin >> a[i];
    for(int i=1;i<n;i++) cin >> b[i];
    for(int i=2;i<n;i++) b[i]+=b[i-1];
    int x=0;
    for(int i=2;i<=n;i++)
    {
        ans[i]=ans[i-1]+a[i-1];
        ans[i]=min(ans[i],c+b[i-1]+x);
        x=min(x,ans[i]-b[i-1]);
    }
    for(int i=1;i<=n;i++)
        cout << ans[i] << (i==n?'\n':' ');
    return 0;
}