首页 > ACM题库 > HDU-杭电 > HDU 3861-The King’s Problem-图-[解题报告]HOJ
2015
04-13

HDU 3861-The King’s Problem-图-[解题报告]HOJ

The King’s Problem

问题描述 :

In the Kingdom of Silence, the king has a new problem. There are N cities in the kingdom and there are M directional roads between the cities. That means that if there is a road from u to v, you can only go from city u to city v, but can’t go from city v to city u. In order to rule his kingdom more effectively, the king want to divide his kingdom into several states, and each city must belong to exactly one state. What’s more, for each pair of city (u, v), if there is one way to go from u to v and go from v to u, (u, v) have to belong to a same state. And the king must insure that in each state we can ether go from u to v or go from v to u between every pair of cities (u, v) without passing any city which belongs to other state.
  Now the king asks for your help, he wants to know the least number of states he have to divide the kingdom into.

输入:

The first line contains a single integer T, the number of test cases. And then followed T cases.

The first line for each case contains two integers n, m(0 < n <= 5000,0 <= m <= 100000), the number of cities and roads in the kingdom. The next m lines each contains two integers u and v (1 <= u, v <= n), indicating that there is a road going from city u to city v.

输出:

The first line contains a single integer T, the number of test cases. And then followed T cases.

The first line for each case contains two integers n, m(0 < n <= 5000,0 <= m <= 100000), the number of cities and roads in the kingdom. The next m lines each contains two integers u and v (1 <= u, v <= n), indicating that there is a road going from city u to city v.

样例输入:

1
3 2
1 2
1 3

样例输出:

2

这题一开始理解有问题。

对于每一个洲,如果洲里面的任意两个城市u,v,如果u有路径到v,则v也要有路径到u。不要求两两城市都存在路径。

用Tarjan求强连通分量缩点,在用二分图求

#ifdef _MSC_VER
#define DEBUG
#endif

#include <fstream>
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <string>
//#include <memory.h>
#include <limits.h>
#include <algorithm>
#include <math.h>
#include <numeric>
#include <functional>
#include <ctype.h>
#include <vector>
using namespace std;

#define  MAX 50005              //题目中可能的最大点数       
int STACK[MAX],top;          //Tarjan 算法中的栈 
bool InStack[MAX];             //检查是否在栈中 
int DFN[MAX];                  //深度优先搜索访问次序 
int Low[MAX];                  //能追溯到的在栈中的最早次序
int ComponentNumber;       //有向图强连通分量个数 
int Index;                 //索引号 
vector <int> Edge[MAX];        //邻接表表示 
int InComponent[MAX];			//记录每个点在第几号强连通分量里
void Tarjan(int i) 
{ 
  int j; 
  DFN[i]=Low[i]=Index++; 
  InStack[i]=true; 
  STACK[++top]=i; 
  for (size_t e=0;e<Edge[i].size();e++) 
  { 
    j=Edge[i][e]; 
    if (DFN[j]==-1) 
    { 
      Tarjan(j); 
      Low[i]=min(Low[i],Low[j]); 
    } 
    else if (InStack[j]) //如果指向的节点j仍在栈中,由于j先于i入栈,则j有到i的通路,同时由于i指向j,则i与j构成回路
      Low[i]=min(Low[i],DFN[j]); 	//如果指向的节点扔在栈中,则指向的节点仍未编入强连通分量
    //如果前面两个判断条件都是错误的话,则i和j不在同一个连通分量中
  } 
  if (DFN[i]==Low[i]) //连通分量中最早进栈的点
  { 
    ComponentNumber++; 
    do 
    { 
      j=STACK[top--]; 
      InStack[j]=false; 
//       Component[ComponentNumber].push_back(j);
      InComponent[j]=ComponentNumber;	//给每一个连通分量上的节点染色
    } 
    while (j!=i); 
  } 
}

void solve(int N)     //N是此图中点的个数,注意是0-indexed! 
{ 
  memset(STACK,-1,sizeof(STACK)); 
  memset(InStack,0,sizeof(InStack)); 
  memset(DFN,-1,sizeof(DFN)); 
  memset(Low,-1,sizeof(Low)); 
  top=0;Index=0;ComponentNumber=0;

  for(int i=1;i<=N;i++) 
    if(DFN[i]==-1) 
      Tarjan(i);    
}

vector<int> v_component_edge[MAX];
int used[MAX];
int mat[MAX];
bool find(const int &x)
{
  size_t i;
  for(i=0;i<v_component_edge[x].size();++i)
  {
    int v=v_component_edge[x][i];
    if(!used[v])
    {
      used[v]=1;
      if(mat[v]==-1 || find(mat[v]))
      {
        mat[v]=x;
        return true;
      }
    }
  }
  return false;
}

int main(void)
{
#ifdef DEBUG  
  freopen("../stdin.txt","r",stdin);
  freopen("../stdout.txt","w",stdout); 
#endif  

  int ncases,n,m,u,v;
  scanf("%d",&ncases);

  while(ncases--)
  {
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i)
    {
      Edge[i].clear();
      v_component_edge[i].clear();
    }
    while(m--)
    {
      scanf("%d%d",&u,&v);
      Edge[u].push_back(v);
    }

    solve(n);

    for(int i=1;i<=n;++i)
      for(size_t j=0;j<Edge[i].size();++j)
        if(InComponent[i]!=InComponent[Edge[i][j]])
          v_component_edge[InComponent[i]].push_back(InComponent[Edge[i][j]]);

    int ans=0;
    memset(mat,-1,sizeof(mat));
    for (int i=1;i<=ComponentNumber;++i)
    {
      memset(used,0,sizeof(used));
      if(find(i)) ++ans;
    }

    printf("%d\n",ComponentNumber-ans);

  }

  return 0;
}

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参考:http://blog.csdn.net/neofung/article/details/7194657