Codeforces Round #234 (Div. 二) D. Dima and Bacteria
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.
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 .
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».
4 4 2 1 3 2 3 0 3 4 0 2 4 1 2 1 2
Yes 0 2 2 0
3 1 2 2 1 1 2 0
Yes 0 -1 -1 0
3 2 2 2 1 1 2 0 2 3 1
Yes 0 1 1 0
3 0 2 1 2
No
题意：
有n个点，m条边，一些点属于一些集合，问集合到集合的距离及是否满足一个集合内的任意两点都存在距离为0的路径。
思路：
最短路+并查集判断是否任意两点都存在距离为0的路径。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 100005
#define MAXN 200005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
ll n,m,k,ans,cnt,tot,flag;
ll c[maxn],sum[maxn],now,oo;
ll dist[505][505];
ll pre[maxn],fa[505];
ll Find(ll x)
{
if(x==pre[x]) return x;
pre[x]=Find(pre[x]);
return pre[x];
}
void Merge(ll x,ll y)
{
ll u=Find(x), v=Find(y);
if(x!=y) pre[u]=v;
}
int main()
{
ll i,j,t,u,v;
while(~scanf("%I64d%I64d%I64d",&n,&m,&k))
{
sum[0]=0;
for(i=1; i<=k; i++)
{
scanf("%I64d",&c[i]);
sum[i]=sum[i-1]+c[i];
}
for(i=1;i<=n;i++) pre[i]=i;
ll u,v,w;
memset(dist,0x3f,sizeof(dist));
oo=dist[0][0];
for(i=1; i<=m; i++)
{
scanf("%I64d%I64d%I64d",&u,&v,&w);
ll posu=lower_bound(sum+1,sum+k+1,u)-sum;
ll posv=lower_bound(sum+1,sum+k+1,v)-sum;
dist[posu][posv]=dist[posv][posu]=min(dist[posu][posv],w);
if(w==0) Merge(u,v);
}
memset(fa,0,sizeof(fa));
flag=1;
for(i=1;i<=n;i++)
{
ll posu=lower_bound(sum+1,sum+k+1,i)-sum;
u=Find(i);
if(fa[posu])
{
if(u!=fa[posu])
{
flag=0;
break ;
}
}
else
{
fa[posu]=u;
}
}
if(flag) printf("Yes\n");
else
{
printf("No\n");
continue ;
}
for(i=1;i<=k;i++) dist[i][i]=0;
for(ll p=1;p<=k;p++)
{
for(i=1;i<=k;i++)
{
for(j=1;j<=k;j++)
{
if(dist[i][j]>dist[i][p]+dist[p][j])
dist[i][j]=dist[i][p]+dist[p][j];
}
}
}
for(i=1; i<=k; i++)
{
for(j=1; j<=k; j++)
{
if(dist[i][j]>=oo) dist[i][j]=-1;
if(j==1) printf("%I64d",dist[i][j]);
else printf(" %I64d",dist[i][j]);
}
printf("\n");
}
}
return 0;
}