TJU2248 Channel Design (最小树形图)

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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 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

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 intusing 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;}