1182. Team Them Up!
Time limit: 1.0 second Memory limit: 64 MB Your task is to divide a number of persons into two teams, in such a way, that:- everyone belongs to one of the teams;
- every team has at least one member;
- every person in the team knows every other person in his team;
- teams are as close in their sizes as possible.
Input
For simplicity, all persons are assigned a unique integer identifier from 1 to N. The first line contains a single integer number N (2 ≤ N ≤ 100) - the total number of persons to divide into teams, followed by N lines - one line per person in ascending order of their identifiers. Each line contains the list of distinct numbers Aij (1 ≤ Aij ≤ N, Aij ≠ i) separated by spaces. The list represents identifiers of persons that ith person knows. The list is terminated by 0.Output
If the solution to the problem does not exist, then write a single message “No solution” (without quotes). Otherwise write a solution on two lines. On the first line write the number of persons in the first team, followed by the identifiers of persons in the first team, placing one space before each identifier. On the second line describe the second team in the same way. You may write teams and identifiers of persons in a team in any order.Sample
| input | output |
|---|---|
5 2 3 5 0 1 4 5 3 0 1 2 5 0 1 2 3 0 4 3 2 1 0 |
3 1 3 5 2 2 4 |
1 /*本题主要是要求补图
2 和dfs求连通块，每个连通块分成两部分x[][],y[][];
3 然后再用01背包求出最小值并标记
4 最后输出计算结果，此处注意x[i][0],y[i][0]是表示i个连通块中两部分的数量
5 v[i]数组用于标记，点i属于哪个组。
6 */
7 #include<iostream>
8 #include<string>
9 #include<cstring>
10 #include<cstdio>
11 #include<cmath>
12 #include<algorithm>
13 using namespace std;
14 const int maxn=110;
15 int e[1001][1001];
16 int v[1001],x[1001][1001],y[1001][1001];
17 int f[1001][1001];
18 int n,m,j,i,p;
19 int tmp;
20 int absw(int x)
21 {
22 return (x<0)?(-x):x;
23 }
24 void init()
25 {
26 cin>>n;
27 memset(e,0,sizeof(e));
28 for(i=0;i<n;i++)//建图
29 {
30 while(cin>>tmp&&tmp)
31 {
32 e[i][tmp-1]=1;
33 }
34 }
35 for(i=0;i<n;i++)//求补图
36 for(j=0;j<i;j++)
37 {
38 if(e[i][j]==1&&e[j][i]==1)
39 e[i][j]=e[j][i]=0;
40 else
41 e[i][j]=e[j][i]=1;
42 }
43 }
44 bool dfs(int k,int p,int xx)//染色、求连通块
45 {
46 for(int it=0;it<n;it++)
47 if(it!=k&&e[it][k]==1)
48 {
49 if(v[it]==-1)
50 {
51 if(xx==1)
52 y[p][++y[p][0]]=it;
53 else
54 x[p][++x[p][0]]=it;
55 v[it]=xx;
56 if(!dfs(it,p,1-xx))return false;
57 }
58 else
59 if(v[k]==v[it])return false;
60
61 }
62 return true;
63 }
64 bool solve()
65 {
66 int ans,g1;
67 p=0;
68 memset(v,-1,sizeof(v));
69 for(i=0;i<n;i++)
70 {
71 if(v[i]==-1)
72 {
73 v[i]=0;x[p][0]=1;x[p][1]=i;y[p][0]=0;
74 if(!dfs(i,p,1)) return false;
75 p++;//p个连通块
76 }
77 }
78 memset(f,255,sizeof(f));
79 f[0][x[0][0]-y[0][0]+maxn]=0;
80 f[0][y[0][0]-x[0][0]+maxn]=1;
81 for(i=1;i<p;i++)//背包问题、f[i][j]表示前i个连通块中当差为j时取x(0)或y(1);
82 for(j=1;j<=maxn*2;j++)
83 {
84 if(f[i-1][j]!=-1)
85 f[i][j+x[i][0]-y[i][0]]=0;
86 if(f[i-1][j]!=-1)
87 f[i][j+y[i][0]-x[i][0]]=1;
88 }
89 ans=-1;
90 for(j=0;j<=maxn*2;j++)
91 if(f[p-1][j]!=-1)
92 if(ans==-1||absw(ans-maxn)>absw(j-maxn))
93 ans=j;//最小差值
94 g1=0;
95 for(i=p-1;i>=0;i--)//背包问题从后往前找
96 if(f[i][ans]==0)
97 {
98 g1+=x[i][0];
99 for(j=1;j<=x[i][0];j++)v[x[i][j]]=1;
100 for(j=1;j<=y[i][0];j++)v[y[i][j]]=2;
101 ans-=(x[i][0]-y[i][0]);
102 }
103 else
104 {
105 g1+=y[i][0];
106 for(j=1;j<=x[i][0];j++)
107 v[x[i][j]]=2;
108 for(j=1;j<=y[i][0];j++)
109 v[y[i][j]]=1;
110 ans-=(y[i][0]-x[i][0]);
111 }
112 //输出计算结果
113 cout<<g1<<' ';
114 for(i=0;i<n;i++)
115 if(v[i]==1)
116 cout<<i+1<<' ';
117 cout<<endl;
118 cout<<n-g1<<' ';
119 for(i=0;i<n;i++)
120 if(v[i]==2)
121 cout<<i+1<<' ';
122 cout<<endl;
123 return true;
124
125 }
126 int main()
127 {
128 init();
129 if(!solve())
130 cout<<"No solution"<<endl;
131 }
View Code
坚持！！！！！
转载于:https://www.cnblogs.com/sdau--codeants/p/3306909.html