TJU2248 Channel Design (最小树形图)

2248.    Channel Design Time Limit: 1.0 Seconds    Memory Limit: 65536K
Total Runs: 1863    Accepted Runs: 600

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

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 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 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;
}