Codeforces 400 D. Dima and Bacteria

判断一个集合内的点能否以花费为0互相抵达。。。。求集合间的最短路

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 

.

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».

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 <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

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<pos+c; j++) be[j]=i;
pos=pos+c;
}

for(int i=0; i<m; i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
if(c==0) Union(a,b);
a=be[a],b=be[b];
g[a][b]=g[b][a]=min(g[b][a],c);
}

int mark=-1,SINE; bool flag=true;
for(int i=1;i<=n;i++)
{
if(be[i]!=mark) SINE=Find(i),mark=be[i];
else if(Find(i)!=SINE) {flag=false; break;};
}

if(!flag) {puts("No");return 0;} else puts("Yes");

for(int l=1; l<=k; l++)
{
for(int i=1; i<=k; i++)
{
for(int j=1; j<=k; j++)
{
g[i][j]=min(g[i][j],g[i][l]+g[l][j]);
}
}
}

for(int i=1; i<=k; i++)
{
for(int j=1; j<=k; j++)
{
int t=(g[i][j]==INF)?-1:g[i][j];
printf("%d ",t);
}
putchar(10);
}

return 0;
}