解題區/解題區 - UVa

UVa 1627 - Team them up

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

現在有 2 個 team，每個 team 至少一個成員，同一個 team 成員之間必然會互相認識，不同 team 的成員之間則不一定。

給定他們互相認識的關係圖，請問他們可以分屬的 team 的情況，並且輸出 team 人數差距最少的那一組。

Sample Input

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
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

1
2
3
4
No solution
3 1 3 5
2 2 4

Solution

對於不認識的兩個人，保證他們在不同的 team。

取補圖，建造相互不認識的兩個人的關係圖，對於每一個 component 做二分圖判定，找到每個二分圖的兩個集合大小。

對於每一個二分圖結果，使用背包問題的解法，將兩個 team 的總數劃分均勻。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
#include <stdio.h> 
#include <string.h>
#include <vector>
#include <algorithm>
#include <assert.h>
using namespace std;
#define MAXN 128
int g[MAXN][MAXN], visited[MAXN];
int color[MAXN], n;
int dfs(int u, int c, vector<int> &s0, vector<int> &s1) {
    visited[u] = 1, color[u] = c;
    if (c == 0)	s0.push_back(u);
    else		s1.push_back(u);
    for (int i = 1; i <= n; i++) {
        if ((!(g[u][i] == 1 && g[i][u] == 1)) && i != u) {
            if (visited[i] == 0) {
                if (!dfs(i, !color[u], s0, s1))
                    return 0;
            } else {
                if (color[u] == color[i])
                    return 0;
            }
        }
    }
    return 1;
}
int main() {
    int testcase, x;
    scanf("%d", &testcase);
    while (testcase--) {
        scanf("%d", &n);
        while (getchar() != '\n');
        memset(g, 0, sizeof(g));
        memset(visited, 0, sizeof(visited));
        memset(color, 0, sizeof(color));
        for (int i = 1; i <= n; i++) {
            while (scanf("%d", &x) == 1 && x)
                g[i][x] = 1;
        }
        int ok = 1, m = 0;
        vector<int> s0[MAXN], s1[MAXN];
        for (int i = 1; i <= n; i++) {
            if (visited[i] == 0) {
                m++;
                ok &= dfs(i, 0, s0[m], s1[m]);
            }
        }
        if (!ok) {
            puts("No solution");
        } else {
            int dp[MAXN][MAXN] = {}, from[MAXN][MAXN] = {};
            dp[0][0] = 1;
            for (int i = 1; i <= m; i++) {
                for (int j = 0; j <= n; j++) {
                    if (dp[i - 1][j]) {
                        dp[i][j + s0[i].size()] = 1;
                        dp[i][j + s1[i].size()] = 1;
                        from[i][j + s0[i].size()] = 0;
                        from[i][j + s1[i].size()] = 1;
                    }
                }
            }
            int ch = -1;
            for (int i = n/2; i >= 1; i--) {
                if (dp[m][i])	ch = i, i = -1;
            }
            assert(ch > 0);
            vector<int> team1, team2;
            for (int i = m; i >= 1; i--) {
                if (from[i][ch] == 0) {
                    ch -= s0[i].size();
                    team1.insert(team1.end(), s0[i].begin(), s0[i].end());
                    team2.insert(team2.end(), s1[i].begin(), s1[i].end());
                } else {
                    ch -= s1[i].size();
                    team1.insert(team1.end(), s1[i].begin(), s1[i].end());
                    team2.insert(team2.end(), s0[i].begin(), s0[i].end());
                }
            }
            printf("%d", team1.size());
            for (int i = 0; i < team1.size(); i++)
                printf(" %d", team1[i]);
            puts("");
            printf("%d", team2.size());
            for (int i = 0; i < team2.size(); i++)
                printf(" %d", team2[i]);
            puts("");
        }
        if (testcase)	puts("");
    }
    return 0;
}
/*
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
*/