19年冬季第二题 PAT甲级 1166 Summit (25分) 与题库1142类似的一般难度题

我会带着你远行 2023-07-02 07:23 13阅读 0赞

汇总贴在博客的置顶：  
2020年3月PAT甲级满分必备刷题技巧

# 题目 #

7-3 Summit (25分)

A summit (峰会) is a meeting of heads of state or government. Arranging the rest areas for the summit is not a simple job. The ideal arrangement of one area is to invite those heads so that everyone is a direct friend of everyone.

Now given a set of tentative arrangements, your job is to tell the organizers whether or not each area is all set.  
Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (≤ 200), the number of heads in the summit, and M, the number of friendship relations. Then M lines follow, each gives a pair of indices of the heads who are friends to each other. The heads are indexed from 1 to N.

Then there is another positive integer K (≤ 100), and K lines of tentative arrangement of rest areas follow, each first gives a positive number L (≤ N), then followed by a sequence of L distinct indices of the heads. All the numbers in a line are separated by a space.  
Output Specification:

For each of the K areas, print in a line your advice in the following format:

*  if in this area everyone is a direct friend of everyone, and no friend is missing (that is, no one else is a direct friend of  
    everyone in this area), print Area X is OK.
 *  if in this area everyone is a direct friend of everyone, yet there are some other heads who may also be invited without breaking the ideal arrangement, print Area X may invite more people, such as H.where H is the smallest index of the head who may be invited.
 *  if in this area the arrangement is not an ideal one, then print Area X needs help. so the host can provide some special service to help the heads get to know each other.

Here X is the index of an area, starting from 1 to K.  
Sample Input:

8 10  
5 6  
7 8  
6 4  
3 6  
4 5  
2 3  
8 2  
2 7  
5 3  
3 4  
6  
4 5 4 3 6  
3 2 8 7  
2 2 3  
1 1  
2 4 6  
3 3 2 1  
Sample Output:

Area 1 is OK.  
Area 2 is OK.  
Area 3 is OK.  
Area 4 is OK.  
Area 5 may invite more people, such as 3.  
Area 6 needs help.

## 这道题没有什么坑，题意如下： ##

不超过200个人，编号从1到N  
先给出朋友关系  
再给出聚会人员编号问是不是全是朋友？  
不是的话就求助；是的话，再分成两种  
一种是还能再请来聚会，就请编号小的  
另一种是请不到人

## 满分代码 ##

#include<iostream>
    #include<vector>
    using namespace std;
    int n,m,k;
    int main(){ 
        scanf("%d %d\n",&n,&m);
        int G[n+5][n+5];
        vector<vector<int>> va(n+5);
        va.clear();
        int u,v,l;
        for(int i=0;i<m;i++){ 
            scanf("%d %d\n",&u,&v);
            G[u][v]=G[v][u]=1;
            va[u].push_back(v);
            va[v].push_back(u);
        }
        scanf("%d\n",&k);
        for(int i=0;i<k;i++){ 
            scanf("%d",&l);
            int a[l];
            set<int> s;
            for(int j=0;j<l;j++){ 
                scanf("%d",&a[j]);
                s.insert(a[j]);
            }
            if(l==1&&va[a[0]].size()==0){ 
                printf("Area %d is OK.\n",i+1);
                continue;
            }
            bool f1=true,f2=true;
            for(int j=0;j<l-1;j++){ 
                for(int k=j+1;k<l;k++){ 
                    int x=a[j],y=a[k];
                    if(G[x][y]!=1){ 
                        f1=false;
                    }
                }
            }
            if(!f1){ 
                printf("Area %d needs help.\n",i+1);
            }else{ 
                int minn=n+1;
                for(int k=1;k<=n;k++){ 
                    bool f3=true;
                    for(auto it:s){ 
                        if(G[k][it]!=1){ 
                            f3=false;
                        }
                    }
                    if(f3 && s.find(k)==s.end() && k<minn){ 
                        minn=k;
                        f2=false;
                        continue;
                    }
                }
    
                if(f2){ 
                    printf("Area %d is OK.\n",i+1);
                }else{ 
                    printf("Area %d may invite more people, such as %d.\n",i+1,minn);
                }
            }
    
        }
        return 0;
    }