首页 > ACM题库 > HDU-杭电 > HDU 3836-Equivalent Sets-图-[解题报告]HOJ
2015
04-13

HDU 3836-Equivalent Sets-图-[解题报告]HOJ

Equivalent Sets

问题描述 :

To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.

输入:

The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.

输出:

The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.

样例输入:

4 0
3 2
1 2
1 3

样例输出:

4
2
Hint
Case 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1.

/*
分析:
    Tarjan+缩点。
    Tarjan求出有几个强连通分量,如果是1的话,那么ans=0;
否则将它们全部连通,连通所需加边max(t1,t2)条,t1为入度
为0的分量数目、t2为出度为0的分量数目。

                                                  2012-10-29
*/

#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#include"stack"
#define N 30000
using namespace std;
int n,m;
int index_s;
int instack[N],DFN[N],LOW[N];
int belong[N],indegree[N],outdegree[N];
struct Eage
{
    int from,to,next;
}eage[2*N];
int tot,head[N];

void add(int a,int b)
{
    eage[tot].from=a;
    eage[tot].to=b;
    eage[tot].next=head[a];
    head[a]=tot++;
}
void getmap()
{
    int i,l;
    int a,b;
    tot=0;
    memset(head,-1,sizeof(head));
    while(m--)  {scanf("%d%d",&a,&b);add(a,b);}
}
stack<int>st;
void Tarjan(int k)
{
    int j,v;
    st.push(k);
    instack[k]=1;
    DFN[k]=LOW[k]=++index_s;
    for(j=head[k];j!=-1;j=eage[j].next)
    {
        v=eage[j].to;
        if(instack[v])  LOW[k]=LOW[k]>DFN[v]?DFN[v]:LOW[k];
        else if(DFN[v]==-1)
        {
            Tarjan(v);
            LOW[k]=LOW[k]>LOW[v]?LOW[v]:LOW[k];
        }
    }
    if(DFN[k]==LOW[k])
    {
        do
        {
            j=st.top();
            st.pop();
            instack[j]=0;
            belong[j]=k;
        }while(j!=k);
    }
}
void getdegree()
{
    int i,l;
    memset(indegree,0,sizeof(indegree));
    memset(outdegree,0,sizeof(outdegree));
    for(i=0;i<tot;i++)
    {
        if(belong[eage[i].from]==belong[eage[i].to])    continue;
        indegree[belong[eage[i].to]]++;
        outdegree[belong[eage[i].from]]++;
    }
}
int main()
{
    int i;
    int temp,t1,t2,ans;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        getmap();

        index_s=0;
        memset(DFN,-1,sizeof(DFN));
        memset(LOW,-1,sizeof(LOW));
        memset(instack,0,sizeof(instack));
        for(i=1;i<=n;i++)   if(DFN[i]==-1) Tarjan(i);

        getdegree();

        temp=t1=t2=0;
        for(i=1;i<=n;i++)
        {
            if(belong[i]!=i)    continue;
            temp++;
            if(indegree[i]==0)  t1++;
            if(outdegree[i]==0) t2++;
        }
        ans=t1>t2?t1:t2;
        if(n<1 || temp==1) ans=0;
        printf("%d\n",ans);
   }
    return 0;
}

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参考:http://blog.csdn.net/ice_crazy/article/details/8123540


,
  1. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”

  2. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。