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Re:为什么枚举边的时候必须为容量为0，我们从源点和汇点dfs分别染色，难道满足条件的边不是两个点颜色不一样就足够了吗？

Posted by 962861784 at 2011-05-30 15:08:43 on Problem 3204
In Reply To:为什么枚举边的时候必须为容量为0，我们从源点和汇点dfs分别染色，难道满足条件的边不是两个点颜色不一样就足够了吗？ Posted by:dut200901102 at 2011-03-02 22:30:19 
> for(int i=0;i<n;++i)
>     {
>         for(int j=box[i];j!=-1;j=edges[j].next)
>         if(!(j&1)&&id[i]+id[edges[j].to]==3)//S能到达的染为1，能到达T的染为2
>         ans++;
>     }
> 这样写WA
> for(int i=0;i<n;++i)
>     {
>         for(int j=box[i];j!=-1;j=edges[j].next)
>         if(!(j&1)&&id[i]+id[edges[j].to]==3&&edges[j].c==0)
>         ans++;
>     }
> 这样写AC
> 我开始质疑我对于割集的理解了，请路过的大牛解疑
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