Problem:
Score : 600 points

Problem Statement

There are N cities in Republic of AtCoder. The size of the i-th city is A_{i}.
Takahashi would like to build N-1 bidirectional roads connecting two cities so that any city can be reached from any other city by using these roads.

Assume that the cost of building a road connecting the i-th city and the j-th city is |i-j| \times D + A_{i} + A_{j}.
For Takahashi, find the minimum possible total cost to achieve the objective.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq D \leq 10^9

1 \leq A_{i} \leq 10^9

A_{i} and D are integers.

Input

Input is given from Standard Input in the following format:

N D
A_1 A_2 ... A_N

Output

Print the minimum possible total cost.

Sample Input 1

3 1
1 100 1

Sample Output 1

106

This cost can be achieved by, for example, building roads connecting City 1, 2 and City 1, 3.

Sample Input 2

3 1000
1 100 1

Sample Output 2

2202

Sample Input 3

6 14
25 171 7 1 17 162

Sample Output 3

497

Sample Input 4

12 5
43 94 27 3 69 99 56 25 8 15 46 8

Sample Output 4

658

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <limits.h>

typedef struct city_t {
    int scale;
    int pos;
    int reachable;
} CITY;

typedef struct road_t {
    int a;
    int b;
} ROAD;

int qsort_compare_road(const void *a, const void *b)
{
    return (*(ROAD *) a).a - (*(ROAD *) b).a;
}

int main(int argc, char *argv[])
{
    CITY city[200000];
    ROAD road[200000 - 1];
    int n, d; // input
    int i, j; // loop counter
    long long minscore, minpos;
    long long score; //
    long long score_sum; //answer
    int n_road;

    scanf("%d %d", &n, &d);

    for (i = 0; i < n; i++) {
        city[i].pos = i;
        scanf("%d", &city[i].scale);
        city[i].reachable = 0;
    }

    //qsort(city, n, sizeof(CITY), qsort_compare_int);

    n_road = 0;
    score_sum = 0;
    for (i = 0; i < n; i++) {
        if (city[i].reachable) {continue;}
        // 探索
        minscore = LLONG_MAX;
        minpos = -1;
        for (j = 0; j < n; j++) {
            if (i == j) {continue;}
            score = (long long) abs(city[i].pos - city[j].pos) * d + city[i].scale + city[j].scale;
            if (score < minscore) {
                minscore = score;
                minpos = j;
            }
        }
        city[i].reachable++;
        city[minpos].reachable++;
        score_sum += minscore;

        if (i < minpos) {
            road[n_road].a = i; //small
            road[n_road].b = minpos;
        } else {
            road[n_road].a = minpos; //small
            road[n_road].b = i;
        }
        n_road++;
    }
    if (n_road != n - 1) {
        city[0].reachable = 1;
        for (i = 1; i < n; i++) {
            city[i].reachable = 0;
        }
    }
    while (n_road != n - 1) {
        // need more road
        qsort(road, n_road, sizeof(ROAD), qsort_compare_road);
        for (i = 0; i < n_road; i++) {
            if (city[road[i].a].reachable) {
                city[road[i].b].reachable = 1;
            } else if (city[road[i].b].reachable) {
                city[road[i].a].reachable = 1;
            }
        }
        // reachableでないのが，孤立した集団
        minscore = LLONG_MAX;
        for (i = 0; i < n; i++) {
            if (!city[i].reachable) {continue;}
            for (j = 0; j < n; j++) {
                if (city[j].reachable) {continue;}
                score = (long long) abs(city[i].pos - city[j].pos) * d + city[i].scale + city[j].scale;
                if (score < minscore) {
                    minscore = score;
                    if (i < j) {
                        road[n_road].a = i;
                        road[n_road].b = j;
                    } else {
                        road[n_road].a = j;
                        road[n_road].b = i;
                    }
                }
            }
        }
        score_sum += minscore;
        n_road++;
    }

    printf("%d\n", score_sum);

    return 0;
}