http://codeforces.com/problemset/problem/400/D

Dima took up the biology of bacteria, as a result of his experiments, he invented k types of bacteria. Overall, there are n bacteria at his laboratory right now, and the number of bacteria of type i equals ci. For convenience, we will assume that all the bacteria are numbered from 1 to n. The bacteria of type ci are numbered from  to .

With the help of special equipment Dima can move energy from some bacteria into some other one. Of course, the use of such equipment is not free. Dima knows m ways to move energy from some bacteria to another one. The way with number i can be described with integers ui, vi and xi mean that this way allows moving energy from bacteria with number ui to bacteria with number vi or vice versa for xi dollars.

Dima's Chef (Inna) calls the type-distribution correct if there is a way (may be non-direct) to move energy from any bacteria of the particular type to any other bacteria of the same type (between any two bacteria of the same type) for zero cost.

As for correct type-distribution the cost of moving the energy depends only on the types of bacteria help Inna to determine is the type-distribution correct? If it is, print the matrix d with size k × k. Cell d[i][j] of this matrix must be equal to the minimal possible cost of energy-moving from bacteria with type i to bacteria with type j.

Input

The first line contains three integers n, m, k (1 ≤ n ≤ 105; 0 ≤ m ≤ 105; 1 ≤ k ≤ 500). The next line contains k integers c1, c2, ..., ck(1 ≤ ci ≤ n). Each of the next m lines contains three integers ui, vi, xi (1 ≤ ui, vi ≤ 105; 0 ≤ xi ≤ 104). It is guaranteed that .

Output

If Dima's type-distribution is correct, print string «Yes», and then k lines: in the i-th line print integers d[i][1], d[i][2], ..., d[i][k](d[i][i] = 0). If there is no way to move energy from bacteria i to bacteria j appropriate d[i][j] must equal to -1. If the type-distribution isn't correct print «No».

 

题意：n个细菌分成k类，求：①是否每一类细菌中的所有细菌都距离为0互相可达②如果前面成立的话，求k类细菌的距离矩阵。

思路：①用并查集把所有的距离0的细菌合并起来就好了。最后看是否每一类细菌都在同一个集合之中。注意未必是一类细菌通过内部路径相互到达，也有可能通过其他类细菌中转来互相可达。

②Floyd跑一遍即可。500^3=1.25e8，本来是交着试一试，但是最后177ms，cf的评测机也太快了，感觉不太合适。

#include<bits/stdc++.h>
using namespace std;
#define maxn 100000+1000

int n,m,k,c[510],s[maxn],p[maxn],d[510][510];
int from[maxn],to[maxn],x[maxn];

int find(int x){return x==p[x]?x:p[x]=find(p[x]);}

bool solve()
{
	for(int i=1;i<=n;i++)p[i]=i;
	for(int i=1;i<=m;i++)if(!x[i])
	{
		int u=from[i],v=to[i];
		if(find(u)!=find(v))p[find(u)]=find(v);
	}	
	for(int pos=1,i=1;i<=k;pos+=c[i],i++)
	{
		for(int j=0;j<c[i];j++)if(find(pos+j)!=find(pos))return false;
	}
	return true;
}

void solve1()
{
	memset(d,0x3f,sizeof(d));
	for(int i=1;i<=k;i++)d[i][i]=0;
	for(int i=1;i<=m;i++)
	{
		int u=s[from[i]],v=s[to[i]];
		if(u!=v)d[u][v]=d[v][u]=min(d[u][v],x[i]);
	}
	for(int l=1;l<=k;l++)
		for(int i=1;i<=k;i++)
			for(int j=1;j<=k;j++){
				d[i][j]=min(d[i][j],d[i][l]+d[l][j]);
			}
				
	printf("Yes\n");
	for(int i=1;i<=k;i++)
	{
		for(int j=1;j<=k;j++)printf("%d ",d[i][j]!=0x3f3f3f3f?d[i][j]:-1);
		puts("");
	}
}

int main()
{
//	freopen("input.in","r",stdin);
	cin>>n>>m>>k;
	for(int i=1;i<=k;i++)cin>>c[i];
	for(int pos=1,i=1;i<=k;pos+=c[i],i++)
	{
		for(int j=0;j<c[i];j++)s[pos+j]=i;
	}	
	for(int i=1;i<=m;i++)scanf("%d%d%d",&from[i],&to[i],&x[i]);
	if(!solve())puts("No");
	else solve1();
	return 0;
}