Question

我部分能够解决这个问题（在一个测试案例中失败了）。我无法弄清楚一个会破坏我的算法的测试用例。

给定图G =（V，E），确定其顶点是否可以分为两组V1和V2，这样V1和V2都在G中形成一个集团。

算法..

a）按照连接顺序对朋友进行排序。（不计算与自我的联系）

b）将表A（TA）和表B（TB）都设置为空。

bool current_placed = false
For (loop) all the friends in the sorted list do the below

  current_placed = false

  if (TA empty) {

      Place current friend on TA

      current_placed = true

  } else {

      // If current friend is connected with all the people already seated on TA.

     if(current friend connected with all on TA){
        Place current friend on TA too.
        current_placed = true
     }
  }

  if(current_placed == false){

     If (TB empty) {

        Place current friend on TB

        current_placed = true

     } else {

       // If current friend is connected with all the people already seated on TB
       if(current friend connected with all on TB){
          Place current friend on TB too.
          current_placed = true;
       } 
    } 
  } 

  if(current_placed == false) {

    ans => NO

    break starting for loop

  } 
// For loop ends

if(current_placed == true)
{
   ans => YES
}

我的代码：

#include<bits/stdc++.h>

using namespace std;

typedef vector<int> VI;
typedef set<int> SI;
typedef vector<VI> VVI;
typedef pair<int,int> PII;
typedef vector<PII> VPII;

bool mat[1004][1004];
int contacts[10004];

bool mycomp(const PII &a, const PII &b){
     return a.second < b.second;
}

int main(){

    int tcs, num_friends, num_pairs;
    int a, b;
    cin >> tcs;

    while(tcs--){

        SI TA, TB;
        cin >> num_friends >> num_pairs;

        memset(mat, 0, sizeof(bool)*1004*1004);
        memset(contacts, 0, sizeof(int)*10004);

        for(int i=0; i<num_pairs; i++){
            cin >> a >> b;
            mat[a][b] = true;
            mat[b][a] = true;
        }

        for(int r=1; r<=num_friends; r++){
            for(int c=1; c<=num_friends; c++){
                if(mat[r][c] and r != c){
                    contacts[r] += 1;
                }
            }
        }

        bool placed = false, connected;

        VPII myvec;
        for(int i=1; i<=num_friends; i++)
            myvec.push_back(PII(i,contacts[i]));

        sort(myvec.begin(), myvec.end(), mycomp);
        VPII :: iterator it;

        for(it = myvec.begin(); it != myvec.end(); ++it){
            PII p = *it;
            //cout << p.first << " " << p.second << endl;
            int i = p.first;    

            placed = false;

            if(TA.empty()){
                TA.insert(i);
                placed = true;
            } else {
                connected = true;
                SI :: iterator itr;
                for(itr = TA.begin(); itr != TA.end(); ++itr){
                    if(mat[i][*itr] == false){
                        connected = false;
                        break;
                    }
                }   
                if(connected == true){
                    TA.insert(i);
                    placed = true;
                }
            }           

            if(placed == false){
                if(TB.empty()){
                    TB.insert(i);
                    placed = true;
                } else {
                    connected = true;
                    SI :: iterator itr;
                    for(itr = TB.begin(); itr != TB.end(); ++itr){
                        if(mat[i][*itr] == false){
                            connected = false;
                            break;
                        }
                    }
                    if(connected == true){
                        TB.insert(i);
                        placed = true;
                    }
                }
            }

            if(placed == false){
                cout << "NO" << endl;
                break;
            }
        }

        if(placed == true) 
            cout << "YES" << endl;
        /*
        SI :: iterator itr;
        cout << endl;
        cout << "SET A" << endl;
        for(itr = TA.begin(); itr != TA.end(); ++itr)
            cout << *itr << " ";
        cout << endl;
        cout << "SET B" << endl;
        for(itr = TB.begin(); itr != TB.end(); ++itr)
            cout << *itr << " ";
        cout << endl; */
    }

return 0;
}

Answer 1

不是暗示，而是反驳。

考虑6个顶点图：

(1, 2), (1, 3), (2, 3)
(4, 5), (5, 6), (4, 6)
(1, 4), (2, 3), (5, 6)

每个人有3个连接。请注意，该图表有2个派系（1, 2, 3和4, 5, 6）。

现在让我们先说一个人1，然后你把他放在桌子A上。一个人4来到第二，你的算法也将他放在表A.之后就不能放置。

我真的没有办法修复这个贪婪的位置。

代码厨师：https：//www.codechef.com/SEPT16/problems/CHFNFRN

1 个答案: