Tju 2248. Channel Design Minimum tree diagram

Minimum tree diagram, test template ....

2248. Channel Designtime limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 2199 Accepted Runs: 740

We need irrigate our farms, but there are only one source of water nearby. So we need build some water channels with minimum cost.

In figure (a), V_{1 indicates the source of water. Other N-1 nodes in the indicate of the farms we need to irrigate. An edge represents you can build a channel between the nodes, to irrigate the target. The integers indicate the cost of a channel between and the nodes.}

Figure (b) represents a design of channels with minimum cost.

Input

there is multiple cases, the first line of all case contains integers  N  and  m   (2≤  N  ≤100; 1≤  m  ≤ 10000),   N  shows the number of nodes. The following  M  lines, each line contains three integers  i j c_ij , means we can build a channel from node  v_i  to node  v_j , which cost  c _IJ . (1≤  i ,   J  ≤  N ;   i  ≠  j ; 1≤  c_ij  ≤100)

The source of water is always V_{1.
The input is terminated by N = M = 0.}

Output

For each case, output a single line contains an integer represents the minimum cost.

If No design can irrigate all the farms, output "impossible" instead.

Sample Input

5 81 2 31 3 52 4 23 1 53 2 53 4 43 5 75 4 33 31 2 31 3 53 2 10 0

Sample Output

176

Problem setter: Hill

Source: tju Contest August 2006 Submit List Runs Forum Statistics

/* ***********************************************author:ckbosscreated time:2015 July 04 Saturday 23:35 05 seconds file Name : tju2248.cpp************************************************ * * #include <iostream> #include <cstdio># Include <cstring> #include <algorithm> #include <string> #include <cmath> #include <cstdlib > #include <vector> #include <queue> #include <set> #include <map>using namespace Std;const int Inf=0x3f3f3f3f;const int maxn=110;int n,m;struct edge{int u,v,cost;}; Edge edge[maxn*maxn];int pre[maxn],id[maxn],vis[maxn],in[maxn];int zhuliu (int root,int n,int M,Edge edge[]) {int res=0, U,v;while (True) {for (int i=0;i<n;i++) in[i]=inf;for (int i=0;i<m;i++) {if (edge[i].u!=edge[i].v&&edge[i ].COST&LT;IN[EDGE[I].V]) {pre[edge[i].v]=edge[i].u;in[edge[i].v]=edge[i].cost;}} for (int i=0;i<n;i++) if (i!=root&&in[i]==inf) Return-1;int Tn=0;memset (id,-1,sizeof (ID)); memset (Vis,-1, sizeof (VIS)); In[root]=0;for (int i=0;i<n;i++) {res+=in[i];v=i;while (vis[v]!=i&&id[v]==-1&&v!=root) {vis[v]=i; v=pre[v];} if (v!=root&&id[v]==-1) {for (int u=pre[v];u!=v;u=pre[u]) id[u]=tn;id[v]=tn++;}} if (tn==0) break;for (int i=0;i<n;i++) if (id[i]==-1) id[i]=tn++;for (int i=0;i<m;) {V=edge[i].v;edge[i].u=id[edge [I].u];edge[i].v=id[edge[i].v];if (EDGE[I].U!=EDGE[I].V) Edge[i++].cost-=in[v];elseswap (Edge[i],edge[--m]);} N=tn;root=id[root];} return res;} int main () {//freopen ("In.txt", "R", stdin),//freopen ("OUT.txt", "w", stdout), while (scanf ("%d%d", &n,&m)!=eof {if (n==0&&m==0) break;for (int i=0;i<m;i++) {int u,v,w;scanf ("%d%d%d", &u,&v,&w); u--; v--; Edge[i].u=u; Edge[i].v=v; Edge[i].cost=w;}        int Ans=zhuliu (0,n,m,edge), if (Ans==-1) puts ("impossible"), Else printf ("%d\n", ans);} return 0;}

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Tju 2248. Channel Design Minimum tree diagram