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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> Codeforces 400 D. Dima and Bacteria

Codeforces 400 D. Dima and Bacteria

編輯:C++入門知識

判斷一個集合內的點能否以花費為0互相抵達。。。。求集合間的最短路

D. Dima and Bacteria time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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 \.<喎?http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">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"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?.

Sample test(s) input
4 4 2
1 3
2 3 0
3 4 0
2 4 1
2 1 2
output
Yes
0 2
2 0
input
3 1 2
2 1
1 2 0
output
Yes
0 -1
-1 0
input
3 2 2
2 1
1 2 0
2 3 1
output
Yes
0 1
1 0
input
3 0 2
1 2
output
No

#include 
#include 
#include 
#include 

using namespace std;

const int INF=0x3f3f3f3f;

int n,m,k,pos=1;
int be[110000];
int g[600][600];

int fa[110000];
void init()
{
    for(int i=1;i<=n;i++) fa[i]=i;
}

int Find(int x)
{
    if(x==fa[x]) return x;
    return fa[x]=Find(fa[x]);
}

void Union(int a,int b)
{
    int A=Find(a),B=Find(b);
    if(A==B) return ;
    fa[A]=B;
}

int main()
{
    scanf("%d%d%d",&n,&m,&k);
    memset(g,63,sizeof(g));
    init();

    for(int i=1; i<=k; i++)
    {
        g[i][i]=0;
        int c;
        scanf("%d",&c);
        for(int j=pos; j


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