Some company is going to hold a fair in Byteland. There are nn towns in Byteland and mm two-way roads between towns. Of course, you can reach any town from any other town using roads.
There are kk types of goods produced in Byteland and every town produces only one type. To hold a fair you have to bring at least ss different types of goods. It costs d(u,v)d(u,v) coins to bring goods from town uu to town vv where d(u,v)d(u,v) is the length of the shortest path from uu to vv . Length of a path is the number of roads in this path.
The organizers will cover all travel expenses but they can choose the towns to bring goods from. Now they want to calculate minimum expenses to hold a fair in each of nn towns.
Input
There are 44 integers nn , mm , kk , ss in the first line of input (1≤n≤1051≤n≤105 , 0≤m≤1050≤m≤105 , 1≤s≤k≤min(n,100)1≤s≤k≤min(n,100) ) — the number of towns, the number of roads, the number of different types of goods, the number of different types of goods necessary to hold a fair.
In the next line there are nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤k1≤ai≤k ), where aiai is the type of goods produced in the ii -th town. It is guaranteed that all integers between 11 and kk occur at least once among integers aiai .
In the next mm lines roads are described. Each road is described by two integers uu vv (1≤u,v≤n1≤u,v≤n , u≠vu≠v ) — the towns connected by this road. It is guaranteed that there is no more than one road between every two towns. It is guaranteed that you can go from any town to any other town via roads.
Output
Print nn numbers, the ii -th of them is the minimum number of coins you need to spend on travel expenses to hold a fair in town ii . Separate numbers with spaces.
Examples
Input
5 5 4 31 2 4 3 21 22 33 44 14 5
Output
2 2 2 2 3
Input
7 6 3 21 2 3 3 2 2 11 22 33 42 55 66 7
Output
1 1 1 2 2 1 1
Note
Let‘s look at the first sample.
To hold a fair in town 11 you can bring goods from towns 11 (00 coins), 22 (11 coin) and 44 (11 coin). Total numbers of coins is 22 .
Town 22 : Goods from towns 22 (00 ), 11 (11 ), 33 (11 ). Sum equals 22 .
Town 33 : Goods from towns 33 (00 ), 22 (11 ), 44 (11 ). Sum equals 22 .
Town 44 : Goods from towns 44 (00 ), 11 (11 ), 55 (11 ). Sum equals 22 .
Town 55 : Goods from towns 55 (00 ), 44 (11 ), 33 (22 ). Sum equals 33 .
Fair
思路：
大体思路就是先储存路线和生产地，再就是用bfs找到每一个产品生产地到其他城市最短距离，最后sort进行排序，
求出前s个得和即可。
bfs
第一次进入while，目的是找到生产x产品所有的生产地，第二次，是对每一个生产地进行往后延生，就是找到与这个地方相邻得
地方，距离再加1.
这个最短路bfs的思路要好好整理，以后还可以用到的
就是用bfs，对它附近的城市距离加+1，这个vector的处理要注意学习！！！！！
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=1e5+1000;
vector<int>exa[maxn];
int mov[maxn][110],n;
void bfs(int x)
{
queue<int>que;
int i,u,v;
que.push(x+n);
while(!que.empty())
{
u=que.front();que.pop();
for(i=0;i<exa[u].size();i++)
{
v=exa[u][i];
if(mov[v][x]==0)
{
mov[v][x]=mov[u][x]+1;
que.push(v);
}
}
}
}
int main()
{
int m,k,s,a,u,v;
scanf("%d %d %d %d",&n,&m,&k,&s);
for(int i=1;i<=n;i++)
{
scanf("%d",&a);
exa[a+n].push_back(i);
}
for(int i=1;i<=m;i++)
{
scanf("%d%d",&u,&v);
exa[u].push_back(v);
exa[v].push_back(u);
}
for(int i=1;i<=k;i++) bfs(i);
for(int i=1;i<=n;i++) sort(mov[i]+1,mov[i]+k+1);
for(int i=1;i<=n;i++)
{
int cnt=0;
for(int j=1;j<=s;j++) cnt+=mov[i][j]-1;
i==n?printf("%d\n",cnt):printf("%d ",cnt);
}
return 0;
}
原文地址：https://www.cnblogs.com/EchoZQN/p/10351643.html