TJU2248 Channel Design (最小树形图)
Total Runs: 1863 Accepted Runs: 600
In Figure (a), V1 indicates the source of water. Other N-1 nodes in the Figure indicate the farms we need to irrigate. An edge represents you can build a channel between the two nodes, to irrigate the target. The integers indicate the cost of a channel between two nodes.
Figure (b) represents a design of channels with minimum cost.
Input
There are multiple cases, the first line of each case contains two 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 cij, means we can build a channel from node Vi to node Vj, which cost cij. (1 ≤ i, j ≤ N; i ≠ j; 1 ≤ cij ≤ 100)
The source of water is always V1.
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 8 1 2 3 1 3 5 2 4 2 3 1 5 3 2 5 3 4 4 3 5 7 5 4 3 3 3 1 2 3 1 3 5 3 2 1 0 0
Sample Output
17 6
Problem setter: Hill
Source: TJU Contest August 2006
题目：http://acm.tju.edu.cn/toj/showp2248.html
分析：典型的最小树形图，直接做就行，由于这题边的长度很小，直接用桶排，算法真正做到O(V^2),刷到Rank 1,激动阿～～～不过之前忘记判断是否存在，wa6次，囧
代码：
#include<cstdio>
using namespace std;
const int mm=22222;
const int mn=222;
struct edge
{
int s,t,w;
}g[mm],h[mm];
int head[mm],next[mm];
int p[mn],q[mn],mark[mn],fp[mn],from[mn],vis[mn],in[mn],w[mn],ans,sum;
int i,j,k,n,m,e,r,mw;
bool huan;
inline void addedge(int u,int v,double c)
{
g[e].s=u,g[e].t=v,g[e].w=c,next[e]=head[u],head[u]=e++;
if(c>mw)mw=c;
}
void dfs(int u)
{
++sum,vis[u]=1;
for(int i=head[u];i>=0;i=next[i])
if(!vis[g[i].t])dfs(g[i].t);
}
inline void init(int& a)
{
char ch=getchar();
while (ch<'0'||ch>'9') ch=getchar();
for (a=0; ch>='0'&&ch<='9'; ch=getchar()) a=a*10+ch-48;
}
void mysort()
{
int i;
for(i=0;i<=n;++i)q[i]=0;
for(i=0;i<=mw;++i)p[i]=0;
for(i=0;i<e;++i)++q[g[i].t],++p[g[i].w];
for(i=1;i<=n;++i)q[i]+=q[i-1];
for(i=n;i>0;--i)q[i]=q[i-1];
q[0]=0;
for(i=1;i<=mw;++i)p[i]+=p[i-1];
for(i=mw;i>0;--i)p[i]=p[i-1];
p[0]=0;
for(i=0;i<e;++i)h[p[g[i].w]++]=g[i];
for(i=0;i<e;++i)g[q[h[i].t]++]=h[i];
}
int main()
{
while(init(n),init(m),n+m)
{
for(i=e=mw=0;i<=n;++i)head[i]=-1;
while(m--)
{
init(i),init(j),init(k);
if(i!=j)addedge(i,j,k);
}
for(sum=i=0;i<=n;++i)vis[i]=0;
dfs(1);
if(sum<n)
{
printf("impossible\n");
continue;
}
mysort();
for(i=0;i<=n;++i)fp[i]=p[i]=-1,in[i]=vis[i]=0,mark[i]=i;
for(i=0;i<e;++i)
if(p[g[i].t]<0)p[g[i].t]=i;
huan=1,ans=sum=0;
while(huan)
{
huan=0;
for(i=2;i<=n;++i)
if(fp[j=mark[i]]>=0)
{
if(fp[i]<0)in[i]+=w[j],mark[i]=mark[mark[i]];
else
{
in[i]+=w[i],ans+=w[i];
if(g[++p[fp[i]]].t!=fp[i])p[fp[i]]=-1;
}
}
for(i=0;i<=n;++i)fp[i]=-1,vis[i]=0;
for(i=2;i<=n;++i)
if(p[i]>=0)
{
if(fp[j=mark[i]]<0||(fp[j]>=0&&w[j]>g[p[i]].w-in[i]))
w[j]=g[p[i]].w-in[i],fp[j]=i,from[j]=mark[g[p[i]].s];
}
for(sum=0,i=2;i<=n;++i)
if(fp[i]>=0)sum+=w[i];
for(i=2;i<=n;++i)
if(!vis[i])
{
r=0,j=i;
while(j>0&&vis[j]>=0)
{
if(vis[j]>0)
{
huan=1;
while(q[--r]!=j)mark[q[r]]=j,vis[q[r]]=-1;
vis[j]=-1;
}
else if(!vis[j])vis[q[r++]=j]=1;
if(fp[j]>=0)j=from[j];
else j=-1;
}
while(r--)vis[q[r]]=fp[q[r]]=-1;
}
}
printf("%d\n",ans+sum);
}
return 0;
}
普通版本：
#include<cstdio>
#define type int
using namespace std;
const int mm=11111;
const int mn=111;
const int oo=1000000000;
int s[mm],t[mm],c[mm];
int id[mn],pre[mn],q[mn],vis[mn];
type in[mn],w[mn];
type Directed_MST(int root,int NV,int NE)
{
type ret=0,sum=0;
int i,j,u,v,r;
bool huan=1;
for(i=0;i<=NV;++i)in[i]=0,id[i]=i,pre[i]=-1;
while(huan)
{
for(i=0;i<=NV;++i)
if(pre[j=id[i]]>=0)
{
if(pre[i]<0)in[i]+=w[j],id[i]=id[j];
else in[i]+=w[i],ret+=w[i];
}
for(i=0;i<=NV;++i)pre[i]=-1,vis[i]=0;
for(i=0;i<NE;++i)
if((u=id[s[i]])!=(v=id[t[i]])&&(w[v]>(j=c[i]-in[t[i]])||pre[v]<0))
pre[v]=u,w[v]=j;
for(i=1;i<=NV;++i)
if(i!=root&&id[i]==i&&pre[i]<0)return -1;
for(pre[root]=-1,sum=i=0;i<=NV;++i)
if(pre[i]>=0)sum+=w[i];
for(huan=i=0;i<=NV;++i)
if(!vis[i])
{
r=0,j=i;
while(j>=0&&vis[j]>=0)
{
if(vis[j]>0)
{
while(q[--r]!=j)id[q[r]]=j,vis[q[r]]=-1;
huan=1,vis[j]=-1;
}
else vis[q[r++]=j]=1,j=pre[j];
}
while(r--)vis[q[r]]=pre[q[r]]=-1;
}
}
return ret+sum;
}
int main()
{
int i,n,m;
while(scanf("%d%d",&n,&m),n+m)
{
for(i=0;i<m;++i)scanf("%d%d%d",&s[i],&t[i],&c[i]);
type ans=Directed_MST(1,n,m);
if(ans<0)printf("impossible\n");
else printf("%d\n",ans);
}
return 0;
}