【CodeForces】-400D 连通图+最短路

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».

Examples
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

 

题意:n个细菌,m个无向边,k个种类 . 输入k种类的个数c[i] sigmac[i]==n,编号按种类来排序.

比如3个种类数量分别为 2 3 4   那么编号1~2属于种类1,3~5属于种类2,6~9属于种类3

同种细菌之间的权值都为0.如果不满足,输出NO.满足,输出多源最短路的距离矩阵

判断不满足:并查集把dis=0的点合并,接下来对c[i]判断

满足的话:输出YES,缩c[i]点为一个点,k<=500,那么O(k^3)的复杂度求出多源最短路,这题不能用bfs,因为权值不相等

#include <bits/stdc++.h>
#define MOD 1000000007
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
 
#define bug1 cout << "bug1" << endl
#define bug2 cout << "bug2" << endl
#define bug3 cout << "bug3" << endl
#define bug4 cout << "bug4" << endl
const int MAX_N=3000+3;
int c[505],par[100005],dis[600][600],ans[600][600],n,m,k;
map<int,int> mp;
 
int Find(int x){
    return x==par[x]? x : par[x]=Find(par[x]) ;
}
 
bool check(){
    int index=1;
    for(int i=1;i<=k;i++){
        int x=Find(index++);
        for(int j=2;j<=c[i];++j){
            if(x!=Find(index++))    return false;
        }
    }
    return true;
}
 
void dijkstra(int st){
    int vis[600];
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=k;i++)   ans[st][i]=dis[st][i];
    ans[st][st]=0;
    vis[st]=1;
    for(int i=1;i<=k-1;i++){
        int mn=INF,index=1;
        for(int j=1;j<=k;j++){
            if(!vis[j] && ans[st][j]<mn){
 
                    index=j,mn=ans[st][j];
            }
        }
        vis[index]=1;
        for(int j=1;j<=k;j++){
            if(!vis[j]) ans[st][j]=min(ans[st][j],ans[st][index]+dis[index][j]);
        }
    }
    return ;
}
 
int main(void){
    cin >> n>> m >>k;
    memset(dis,INF,sizeof(dis));
    for(int i=1;i<=n;i++)   par[i]=i;
    for(int i=1;i<=k;i++)   scanf("%d",&c[i]);
    int index=1;
    for(int i=1;i<=k;i++){
        for(int j=1;j<=c[i];j++){
            mp[index++]=i;
        }
    }
    for(int i=1;i<=m;i++){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        if(!w){
            int a=Find(u);
            int b=Find(v);
            par[a]=b;
            dis[mp[u]][mp[v]]=0,dis[mp[v]][mp[u]]=0;
        }
        else{
            dis[mp[u]][mp[v]]=min(dis[mp[u]][mp[v]],w);
            dis[mp[v]][mp[u]]=min(dis[mp[v]][mp[u]],w);
        }
    }
    if(!check()){
        cout <<"No"<<endl;
        return 0;
    }
    memset(ans,INF,sizeof(ans));
    for(int i=1;i<=k;i++)   dijkstra(i);
    cout <<"Yes" <<endl;
    for(int i=1;i<=k;i++){
        for(int j=1;j<=k;j++){
            if(ans[i][j]==INF)   cout <<"-1 ";
            else printf("%d ",ans[i][j]);
        }
        puts("");
    }
    return 0;
}