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
好吧，cf第四道居然这么水，虽然死长死长的，每个类型中取一个点深搜只走权为0的路，然后判断能不能覆盖同类型的所有点即可判断，然后更新dis[i][j] ( 1<=i<= k, 1<=j<=k)的值，走一次floyd即可
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 6 using namespace std; 7 8 #define maxn 100005 9 10 int n,m,k; 11 int c[505],first[maxn],next[2 * maxn],v[2 * maxn],x[2 * maxn],dis[505][505],u1[2 * maxn]; 12 bool flag = 1; 13 bool vis[maxn]; 14 15 void addedge(int a,int b,int id) { 16 int e = first[a]; 17 next[id] = e; 18 first[a] = id; 19 } 20 21 void dfs(int u,int type) { 22 vis[u] = 1; 23 //printf("u = %d type = %d\n",u,type); 24 for(int e = first[u]; e != -1; e = next[e]) { 25 if(!vis[ v[e] ] && x[e] == 0) { 26 dfs(v[e],type); 27 } 28 } 29 } 30 void solve() { 31 for(int i = 1; i <= k; i++) { 32 memset(vis,0,sizeof(vis)); 33 dfs(c[i],i); 34 35 for(int j = c[i - 1] + 1; j <= c[i]; j++) { 36 if(!vis[j]) { 37 flag = 0; 38 //printf(" i = %d\n",i); 39 return; 40 } 41 } 42 43 } 44 } 45 46 void floyd() { 47 for(int p = 1; p <= k; p++) { 48 for(int i = 1; i <= k; i++) { 49 for(int j = 1; j <= k; j++) { 50 if(dis[i][p] != -1 && dis[p][j] != -1) { 51 dis[i][j] = dis[i][j] == -1 ? dis[i][p] + dis[p][j] : 52 min(dis[i][j],dis[i][p] + dis[p][j]); 53 } 54 } 55 } 56 } 57 } 58 void output() { 59 if(flag) { 60 printf("Yes\n"); 61 62 for(int i = 0; i < 2 * m; i += 2 ) { 63 int id1,id2; 64 id1 = lower_bound(c + 1,c + k + 1,u1[i]) - c; 65 id2 = lower_bound(c + 1,c + k + 1,v[i]) - c; 66 if(id1 == id2) continue; 67 dis[id1][id2] = dis[id2][id1] = dis[id1][id2] == -1 ? 68 x[i] : min(dis[id1][id2],x[i]); 69 } 70 71 floyd(); 72 73 for(int i = 1; i <= k; i++) { 74 for(int j = 1; j <= k; j++) { 75 printf("%d",dis[i][j]); 76 if(j != k) printf(" "); 77 } 78 printf("\n"); 79 } 80 } else { 81 printf("No\n"); 82 } 83 } 84 85 86 int main() { 87 88 freopen("sw.in","r",stdin); 89 90 scanf("%d%d%d",&n,&m,&k); 91 92 for(int i = 1; i <= k; i++) { 93 scanf("%d",&c[i]); 94 } 95 96 for(int i = 1; i <= k; i++) { 97 c[i] += c[i - 1]; 98 } 99 100 for(int i = 1; i <= k; i++) { 101 for(int j = 1; j <= k; j++) { 102 if(i == j) dis[i][j] = 0; 103 else dis[i][j] = -1; 104 } 105 } 106 107 108 for(int i = 1; i <= n; i++) first[i] = -1; 109 110 for(int i = 0; i < 2 * m; i = i + 2) { 111 int a,b,w; 112 scanf("%d%d%d",&u1[i],&v[i],&x[i]); 113 v[i + 1] = u1[i]; 114 u1[i + 1] = v[i]; 115 x[i + 1] = x[i]; 116 addedge(u1[i],v[i],i); 117 addedge(v[i],u1[i],i + 1); 118 } 119 120 solve(); 121 122 output(); 123 124 return 0; 125 126 127 }
原文：http://www.cnblogs.com/hyxsolitude/p/3585235.html
