Ultraviolet A 10600 and minor tree Creation

This question involves the next generation of a tree. It is necessary to first understand what the next generation of a tree is.

The second generation of tree, Gu mingzhiyi. One Theorem is that the secondary spanning tree can be obtained from the smallest Spanning Tree exchanging an edge. How can this prove that you can search for it online. However, it is important to note that the exchanged edge must be unused (that is, not the edge of the minimal spanning tree ). In addition, the edge to be switched must exist and be the edge of the maximum edge weight of the shortest path between two points.

Therefore, the edge of the maximum edge weight can be obtained when the minimum spanning tree is obtained.

1 void prim () {2 memset (F, 0, sizeof (f); 3 least [1] = 0; 4 for (INT I = 2; I <= N; I ++) {5 least [I] = map [1] [I]; 6 if (least [I]! = Inf) 7 pre [I] = 1; 8} 9 vis [1] = true; 10 for (INT I = 1; I <= N; I ++) {11 int min = inf, P =-1; 12 for (int K = 1; k <= N; k ++) {13 if (least [k] <min &&! Vis [k]) {14 p = K; min = least [k]; 15} 16} 17 18 if (P =-1) return; 19 ans1 + = min; 20 used [pre [p] [p] = used [p] [pre [p] = 1; 21 for (Int J = 1; j <= N; j ++) {// It should be the accessed vertex, so that the edge is 22 if (vis [J]) 23 f [J] [p] = max (F [J] [pre [p], map [pre [p] [p]); 24 f [p] [J] = f [J] [p]; 25} 26 vis [p] = true; 27 for (int K = 1; k <= N; k ++) {28 If (! Vis [k] & map [p] [k] <least [k]) {29 least [k] = map [p] [k]; 30 pre [k] = P; 31} 32} 33} 34}

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Uv10600 is a bare sub-tree question.

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5  6 using namespace std; 7 const int MAXN=105; 8 const int inf=10000000; 9 int map[MAXN][MAXN]; int f[MAXN][MAXN];10 bool exist[MAXN][MAXN],used[MAXN][MAXN];11 int least[MAXN],pre[MAXN]; bool vis[MAXN]; 12 int n,m,ans1,ans2;13 14 void prim(){15 memset(f,0,sizeof(f));16 least[1]=0;17 for(int i=2;i<=n;i++){18 least[i]=map[1][i];19 if(least[i]!=inf)20 pre[i]=1;21 }22 vis[1]=true;23 for(int i=1;i<=n;i++){24 int min=inf,p=-1;25 for(int k=1;k<=n;k++){26 if(least[k]<min&&!vis[k]){27 p=k; min=least[k];28 }29 }30 31 if(p==-1) return ;32 ans1+=min;33 used[pre[p]][p]=used[p][pre[p]]=1;34 for(int j=1;j<=n;j++){35 if(vis[j])36 f[j][p]=max(f[j][pre[p]],map[pre[p]][p]);37 f[p][j]=f[j][p];38 }39 vis[p]=true;40 for(int k=1;k<=n;k++){41 if(!vis[k]&&map[p][k]<least[k]){42 least[k]=map[p][k];43 pre[k]=p;44 }45 }46 }47 }48 49 void secondT(){50 ans2=inf;51 for(int i=1;i<=n;i++){52 for(int j=1;j<=n;j++){53 if(exist[i][j]&&!used[i][j]&&ans1 + map[i][j] - f[i][j] < ans2)54 ans2=ans1 + map[i][j] - f[i][j];55 }56 }57 }58 59 int main(){60 int T; 61 scanf("%d",&T);62 while(T--){63 scanf("%d%d",&n,&m);64 int u,v,d;65 ans1=ans2=0;66 memset(exist,-1,sizeof(exist));67 for(int i=1;i<=n;i++){68 for(int j=i;j<=n;j++)69 map[i][j]=map[j][i]=inf;70 }71 for(int i=1;i<=m;i++){72 scanf("%d%d%d",&u,&v,&d);73 map[u][v]=map[v][u]=d;74 exist[u][v]=exist[v][u]=1;75 }76 memset(pre,-1,sizeof(pre));77 memset(used,0,sizeof(used));78 memset(vis,false,sizeof(vis));79 prim();80 secondT();81 printf("%d %d\n",ans1,ans2);82 }83 return 0;84 }

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