VK Cup 2017 - Round 3 D. Perishable Roads(最短路)

In the country of Never, there are n cities and a well-developed road system. There is exactly one bidirectional road between every pair of cities, thus, there are as many as  roads! No two roads intersect, and no road passes through intermediate cities. The art of building tunnels and bridges has been mastered by Neverians.

An independent committee has evaluated each road of Never with a positive integer called the perishability of the road. The lower the road's perishability is, the more pleasant it is to drive through this road.

It's the year of transport in Never. It has been decided to build a museum of transport in one of the cities, and to set a single signpost directing to some city (not necessarily the one with the museum) in each of the other cities. The signposts must satisfy the following important condition: if any Neverian living in a city without the museum starts travelling from that city following the directions of the signposts, then this person will eventually arrive in the city with the museum.

Neverians are incredibly positive-minded. If a Neverian travels by a route consisting of several roads, he considers the perishability of the route to be equal to the smallest perishability of all the roads in this route.

The government of Never has not yet decided where to build the museum, so they consider all n possible options. The most important is the sum of perishabilities of the routes to the museum city from all the other cities of Never, if the travelers strictly follow the directions of the signposts. The government of Never cares about their citizens, so they want to set the signposts in a way which minimizes this sum. Help them determine the minimum possible sum for all n possible options of the city where the museum can be built.

Input

The first line contains a single integer n (2 ≤ n ≤ 2000) — the number of cities in Never.

The following n - 1 lines contain the description of the road network. The i-th of these lines contains n - i integers. The j-th integer in thei-th line denotes the perishability of the road between cities i and i + j.

All road perishabilities are between 1 and 109, inclusive.

Output

For each city in order from 1 to n, output the minimum possible sum of perishabilities of the routes to this city from all the other cities of Never if the signposts are set in a way which minimizes this sum.

Examples

input

3
1 2
3

output

2
2
3

input

6
2 9 9 6 6
7 1 9 10
9 2 5
4 10
8

output

6
5
7
5
7
11

Note

The first example is explained by the picture below. From left to right, there is the initial road network and the optimal directions of the signposts in case the museum is built in city 1, 2 and 3, respectively. The museum city is represented by a blue circle, the directions of the signposts are represented by green arrows.

For instance, if the museum is built in city 3, then the signpost in city 1 must be directed to city 3, while the signpost in city 2 must be directed to city 1. Then the route from city 1 to city 3 will have perishability 2, while the route from city 2 to city 3 will have perishability 1. The sum of perishabilities of these routes is 3.

题意:给定一个n个点的带边权完全图(n <= 2000),规定一棵有根生成树的花费为树上每个点到根的边权的最小值之和,问以每个点为根的生成树的最小花费是多少.
分析:大概可以发现每个点为根的最优解一定是一条链,然后边权最小的边一定在最优解中,这样我们可以先把所有边的边权先减去min(edg),然后我们把一个最优解的边按照边权排序,设权为0的边是第k个,那么前k-2条边的边权一定是递减的,因为如果出现了递增的情况那么我们直接把这条边去掉然后替换成一条到权值为0的边的一端点的边结果一定不会更差,这样就有一个n^3的算法,先求出任意两点之间的最短路,根据前边的结论第i个点为根的答案就是min(dis[i][j] + f[j]*2),其中f[j]表示从第j个点出发的权值最小的边的权值,这个算法通过一个特殊的操作可以优化到n^2,我们把原图重构,添加一个虚点,刚开始虚点到每个点的距离为2*f[j],然后对这个图直接跑一边最短路,最后d[i]+(n-1)*min就是每个点的答案.

#include 
#define N 2005
#define MOD 1000000007
#define INF 114748364700000ll
using namespace std;
typedef long long ll;
int n;
ll Min,g[N][N],f[N];
bool vis[N];
int main()
{
    memset(f,0x3f,sizeof(f));
    Min = INF;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
     for(int j = i+1;j <= n;j++)
     {
        scanf("%I64d",&g[i][j]);
        g[j][i] = g[i][j];
        if(Min > g[i][j]) Min = g[i][j];
     }
    for(int i = 1;i <= n;i++)
     for(int j = 1;j <= n;j++)
      if(i != j)
      {
          g[i][j] -= Min;
          f[i] = min(f[i],2*g[i][j]);
      }
    for(int i = 1;i <= n;i++)
    {
        int t = 0;
        for(int j = 1;j <= n;j++)
         if(f[t] > f[j] && !vis[j]) t = j;
        vis[t] = 1;
        for(int j = 1;j <= n;j++) f[j] = min(f[j],f[t] + g[t][j]);
    }
    for(int i = 1;i <= n;i++) cout<