D. Fair(Codeforces Round #485 (Div. 2))

D. Fair
time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output

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 ssdifferent 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 uuto 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 nnmmkkss in the first line of input (1n1051≤n≤1050m1050≤m≤1051skmin(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 (1aik1≤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 (1u,vn1≤u,v≤nuvu≠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
Copy
5 5 4 3
1 2 4 3 2
1 2
2 3
3 4
4 1
4 5
output
Copy
2 2 2 2 3 
input
Copy
7 6 3 2
1 2 3 3 2 2 1
1 2
2 3
3 4
2 5
5 6
6 7
output
Copy
1 1 1 2 2 1 1 

思路:多源最短路

AC:代码

时间:960ms

#include <bits/stdc++.h>


using namespace std;

const int maxn = 1e5+5;
#define INF  0x3f3f3f3f

int good[maxn],path[maxn][105],vis[maxn];
int n,m,s,k;

vector<int>G[maxn];
vector<int>E[120];

struct node{
	int x,s;
	node(int _x,int _s)
	{
		x = _x;
		s = _s;
	}
};

void bfs(int st)
{
	queue<node>que;
	for(int i = 0;i<E[st].size();i++){
		int e = E[st][i];
		que.push(node(e,0));
		path[E[st][i]][st] = 0;
		vis[E[st][i]] = 1;
	}
	while(!que.empty())
	{
		node q = que.front();que.pop();
		path[q.x][st] = min(q.s,path[q.x][st]);
		for(int i = 0;i<G[q.x].size();i++){
			int e = G[q.x][i];
			if(vis[e]) continue;
			vis[e] = 1;
			que.push(node(e,q.s+1));
		}
	}
}

int main()
{
	int a,b;
	scanf("%d %d %d %d",&n,&m,&s,&k);
	for (int i = 1;i<=n;i++){
		scanf("%d",&good[i]);
		E[good[i]].push_back(i);
	}
	for (int i = 1;i<=m;i++){
		scanf("%d %d",&a,&b);
		G[a].push_back(b);
		G[b].push_back(a);
	}
	memset(path,INF,sizeof(path));
	for(int i = 1;i<=s;i++){
		memset(vis,0,sizeof(vis));
		bfs(i);
	}
	long long sum;
	for (int i = 1;i<=n;i++){
		sum = 0;
		sort(path[i]+1,path[i]+s+1);
		for(int j = 1;j<=k;j++){
			sum += path[i][j];
		}
		printf("%lld ",sum);
	}
	return 0;
}

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转载自blog.csdn.net/acer12138/article/details/80580121

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