codeforces 400 D Dima and Bacteria

codeforces 400 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.

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：给定一些细菌的群落 然后每个群落里都有细菌 求 同一个群落里的细菌能否 不花费代价到达

如果不行输出NO可以输出YES 然后还需要求 各个细菌群落之间的最短路

首先并查集 判断是否每个群落细菌都能不花费代价到达

然后缩点跑floyd即可

#include#include#include#define inf 0x3f3f3f3f#define N 110000using namespace std;inline char gc(){ static char now[1<<16],*S,*T; if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;} return *S++;}inline int read(){ int x=0;char ch=gc(); while (ch<'0'||ch>'9') ch=gc(); while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();} return x;}int mp[550][550],fa[N],c[550],mark[N],n,m,k;inline int find(int x){ return fa[x]==x?x:fa[x]=find(fa[x]);}int main(){// freopen("cf.in","r",stdin); n=read();m=read();k=read();int last=0; for (int i=1;i<=k;++i) { c[i]=read();for (int j=last+1;j<=last+c[i];++j) mark[j]=i;last=last+c[i]; } for (int i=1;i<=n;++i) fa[i]=i;memset(mp,0x3f,sizeof(mp)); for (int i=1;i<=k;++i) mp[i][i]=0; for (int i=1;i<=m;++i){ int x=read(),y=read(),z=read(); if (z==0){ int xx=find(x),yy=find(y);fa[xx]=yy; } mp[mark[x]][mark[y]]=min(z,mp[mark[x]][mark[y]]); mp[mark[y]][mark[x]]=min(z,mp[mark[y]][mark[x]]); }int now=0;int fa1=0; for (int i=1;i<=n;++i){ if (mark[i]!=now){ fa1=find(i);now=mark[i];continue; } if (find(i)!=fa1){ printf("No");return 0; } } printf("Yes\n"); for (int kk=1;kk<=k;++kk){ for (int i=1;i<=k;++i){ for (int j=1;j<=k;++j){ mp[i][j]=min(mp[i][j],mp[i][kk]+mp[kk][j]); } } } for (int i=1;i<=k;++i){ for (int j=1;j<=k;++j){ if (mp[i][j]==inf) printf("-1 ");else printf("%d,mp[i][j]); }printf("\n"); } return 0;}

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