题目链接:Dima and Bacteria
Dima took up the biology of bacteria, as a result of his experiments, he invented ktypes 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 mlines 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个点，m条边，每条边是双向的，并且有权值。n个点分为k个种类，每个种类有Ci个点
如果任意两个相同点之间转移，可以找到一条路径，该路径满足经过的边的权值和为0则输出Yes，然后输出任意两个种类的点之间转移的最小路径。否则输出No；
分析：
两个问题：
1.第一个问题，相当于判断同类型的是否在一个集合内，并查集
2.第二个问题，数据范围不大，直接floy搞就行
细节处理较多，具体看代码把
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;const int N=1e5;
const int M=505;
int f[N],c[M],d[M][M],clas[N];
const int INF = 0x3f3f3f3f;
int n,m,q;
int getfa(int x) {return x == f[x] ? x : f[x] = getfa(f[x]);
}bool judge()
{int p=1;for(int i=1;i<=q;i++) ///利用种类的标号是连续的{int u=getfa(p);for(int j=0;j<c[i];j++){int v=getfa(p);if(u!=v) return 0;p++;}}return 1;
}
void Floyd() {for (int i = 1; i <= q; i++) d[i][i] = 0;for (int x = 1; x <= q; x++) {for (int i = 1; i <= q; i++) {for (int j = 1; j <= q; j++) {d[i][j] = min(d[i][j], d[i][x] + d[x][j]);}}}
}int main()
{scanf("%d%d%d",&n,&m,&q);int cnt=1;for(int i=1;i<=n;i++)f[i]=i;memset(d,INF,sizeof(d));for(int i=1;i<=q;i++){scanf("%d",&c[i]);for(int j=0;j<c[i];j++){clas[cnt++]=i;}}for(int i=1;i<=m;i++){int u,v,w;scanf("%d%d%d",&u,&v,&w);if(w==0){int x=getfa(u);int y=getfa(v);if(x!=y)f[y]=x;}d[clas[u]][clas[v]]=d[clas[v]][clas[u]]=min(d[clas[u]][clas[v]],w);}if(judge()){printf("Yes\n");Floyd();for(int i = 1; i <= q; i++) {for (int j = 1; j < q; j++) printf("%d ", d[i][j] != INF ? d[i][j] : -1);printf("%d\n", d[i][q] != INF ? d[i][q] : -1);}}elseprintf("No\n");return 0;
}