UVA1627-Team them up!（二分图判断+动态规划）

Problem UVA1627-Team them up!

Total Submissions:1228  Solved:139

Time Limit: 3000 mSec

Problem Description

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.

This task may have many solutions. You are to find and output any solution, or to report that the solution does not exist.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs. For simplicity, all persons are assigned a unique integer identifier from 1 to N. Thefirstlineintheinputfilecontainsasingleintegernumber N (2 ≤ N ≤ 100) —thetotalnumber 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 i-th person knows. The list is terminated by ‘0’.

 Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line. If the solution to the problem does not exist, then write a single message ‘No solution’ (without quotes) to the output file. Otherwise write a solution on two lines. On the first line of the output file 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

2
 
5
3 4 5 0
1 3 5 0
2 1 4 5 0
2 3 5 0
1 2 3 4 0
 
5
2 3 5 0
1 4 5 3 0
1 2 5 0
1 2 3 0
4 3 2 1 0

 

Sample Output

No solution

3 1 3 5
2 2 4

题解：有一阵子没写博客了，感到很内疚，这个东西还是要坚持。这个题用的东西比较杂，但每一部分都不算难，首先是二分图的判断，染色法dfs很好做，之后就是一个01背包的变形，不过做法似乎和背包没什么关系，数据小，比较暴力的方式就能过，最后是输出解，用的是lrj之前讲的方法。

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn =  + ;

 int n, tot, belong[maxn];
 bool gra[maxn][maxn];
 vector<int> team[maxn][];
 int delta[maxn];

 bool dfs(int u,int flag) {
     belong[u] = flag;
     team[tot][flag - ].push_back(u);
     for (int v = ; v <= n; v++) {
         if (gra[u][v] && u != v) {
             if (belong[v] == flag) return false;
             if (!belong[v] && !dfs(v,  - flag)) return false;
         }
     }
     return true;
 }

 bool build_graph() {
     memset(belong, , sizeof(belong));
     for (int u = ; u <= n; u++) {
         if (!belong[u]) {
             tot++;
             team[tot][].clear();
             team[tot][].clear();
             if (!dfs(u, )) return false;
         }
     }
     return true;
 }

 bool dp[maxn][maxn << ];
 vector<int> ans, ans1;

 void DP() {
     for (int i = ; i <= tot; i++) {
         delta[i] = team[i][].size() - team[i][].size();
         //printf("(%d-%d) = %d\n", team[i][0].size(), team[i][1].size(), delta[i]);
     }
     memset(dp, false, sizeof(dp));
     dp[][ + n] = true;
     for (int i = ; i <= tot; i++) {
         for (int j = -n; j <= n; j++) {
             if (dp[i][j + n]) dp[i + ][j + n + delta[i + ]] = dp[i + ][j + n - delta[i + ]] = true;
         }
     }

     int res = ;
     for (int j = ; j <= n; j++) {
         if (dp[tot][n + j]) {
             res = n + j;
             break;
         }
         if (dp[tot][n - j]) {
             res = n - j;
             break;
         }
     }
     ans.clear(), ans1.clear();
     for (int i = tot; i >= ; i--) {
         if (dp[i - ][res - delta[i]]) {
             for (int j = ; j < (int)team[i][].size(); j++) {
                 ans.push_back(team[i][][j]);
             }
             for (int j = ; j < (int)team[i][].size(); j++) {
                 ans1.push_back(team[i][][j]);
             }
             res -= delta[i];
         }
         else if (dp[i - ][res + delta[i]]) {
             for (int j = ; j < (int)team[i][].size(); j++) {
                 ans1.push_back(team[i][][j]);
             }
             for (int j = ; j < (int)team[i][].size(); j++) {
                 ans.push_back(team[i][][j]);
             }
             res += delta[i];
         }
     }
     printf("%d", ans.size());
     for (int i = ; i < (int)ans.size(); i++) printf(" %d", ans[i]);
     printf("\n");
     printf("%d", ans1.size());
     for (int i = ; i < (int)ans1.size(); i++) printf(" %d", ans1[i]);
     printf("\n");
 }

 int main()
 {
     //freopen("input.txt", "r", stdin);
     int iCase;
     scanf("%d", &iCase);
     while (iCase--) {
         tot = ;
         memset(gra, false, sizeof(gra));
         scanf("%d", &n);
         int v;
         for (int u = ; u <= n; u++) {
             while (true) {
                 scanf("%d", &v);
                 if (v == ) break;
                 gra[u][v] = true;
             }
         }
         for (int u = ; u <= n; u++) {
             for (int v = ; v < u; v++) {
                 if (!gra[u][v] || !gra[v][u]) gra[u][v] = gra[v][u] = true;
                 else gra[u][v] = gra[v][u] = false;
             }
         }

         if (n ==  || !build_graph()) printf("No solution\n");
         else DP();

         if (iCase) printf("\n");
     }
     return ;
 }