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

 

題解：有一陣子沒寫部落格了，感到很內疚，這個東西還是要堅持。這個題用的東西比較雜，但每一部分都不算難，首先是二分圖的判斷，染色法dfs很好做，之後就是一個01揹包的變形，不過做法似乎和揹包沒什麼關係，資料小，比較暴力的方式就能過，最後是輸出解，用的是lrj之前講的方法。

 

  1 #include <bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 const int maxn = 100 + 5;
  6 
  7 int n, tot, belong[maxn];
  8 bool gra[maxn][maxn];
  9 vector<int> team[maxn][2];
 10 int delta[maxn];
 11 
 12 bool dfs(int u,int flag) {
 13     belong[u] = flag;
 14     team[tot][flag - 1].push_back(u);
 15     for (int v = 1; v <= n; v++) {
 16         if (gra[u][v] && u != v) {
 17             if (belong[v] == flag) return false;
 18             if (!belong[v] && !dfs(v, 3 - flag)) return false;
 19         }
 20     }
 21     return true;
 22 }
 23 
 24 bool build_graph() {
 25     memset(belong, 0, sizeof(belong));
 26     for (int u = 1; u <= n; u++) {
 27         if (!belong[u]) {
 28             tot++;
 29             team[tot][0].clear();
 30             team[tot][1].clear();
 31             if (!dfs(u, 1)) return false;
 32         }
 33     }
 34     return true;
 35 }
 36 
 37 bool dp[maxn][maxn << 2];
 38 vector<int> ans, ans1;
 39 
 40 void DP() {
 41     for (int i = 1; i <= tot; i++) {
 42         delta[i] = team[i][0].size() - team[i][1].size();
 43         //printf("(%d-%d) = %d\n", team[i][0].size(), team[i][1].size(), delta[i]);
 44     }
 45     memset(dp, false, sizeof(dp));
 46     dp[0][0 + n] = true;
 47     for (int i = 0; i <= tot; i++) {
 48         for (int j = -n; j <= n; j++) {
 49             if (dp[i][j + n]) dp[i + 1][j + n + delta[i + 1]] = dp[i + 1][j + n - delta[i + 1]] = true;
 50         }
 51     }
 52 
 53     int res = 0;
 54     for (int j = 0; j <= n; j++) {
 55         if (dp[tot][n + j]) {
 56             res = n + j;
 57             break;
 58         }
 59         if (dp[tot][n - j]) {
 60             res = n - j;
 61             break;
 62         }
 63     }
 64     ans.clear(), ans1.clear();
 65     for (int i = tot; i >= 1; i--) {
 66         if (dp[i - 1][res - delta[i]]) {
 67             for (int j = 0; j < (int)team[i][0].size(); j++) {
 68                 ans.push_back(team[i][0][j]);
 69             }
 70             for (int j = 0; j < (int)team[i][1].size(); j++) {
 71                 ans1.push_back(team[i][1][j]);
 72             }
 73             res -= delta[i];
 74         }
 75         else if (dp[i - 1][res + delta[i]]) {
 76             for (int j = 0; j < (int)team[i][0].size(); j++) {
 77                 ans1.push_back(team[i][0][j]);
 78             }
 79             for (int j = 0; j < (int)team[i][1].size(); j++) {
 80                 ans.push_back(team[i][1][j]);
 81             }
 82             res += delta[i];
 83         }
 84     }
 85     printf("%d", ans.size());
 86     for (int i = 0; i < (int)ans.size(); i++) printf(" %d", ans[i]);
 87     printf("\n");
 88     printf("%d", ans1.size());
 89     for (int i = 0; i < (int)ans1.size(); i++) printf(" %d", ans1[i]);
 90     printf("\n");
 91 }
 92 
 93 int main()
 94 {
 95     //freopen("input.txt", "r", stdin);
 96     int iCase;
 97     scanf("%d", &iCase);
 98     while (iCase--) {
 99         tot = 0;
100         memset(gra, false, sizeof(gra));
101         scanf("%d", &n);
102         int v;
103         for (int u = 1; u <= n; u++) {
104             while (true) {
105                 scanf("%d", &v);
106                 if (v == 0) break;
107                 gra[u][v] = true;
108             }
109         }
110         for (int u = 1; u <= n; u++) {
111             for (int v = 1; v < u; v++) {
112                 if (!gra[u][v] || !gra[v][u]) gra[u][v] = gra[v][u] = true;
113                 else gra[u][v] = gra[v][u] = false;
114             }
115         }
116 
117         if (n == 1 || !build_graph()) printf("No solution\n");
118         else DP();
119 
120         if (iCase) printf("\n");
121     }
122     return 0;
123 }