题目传送门
1 /*
2 题意：三种人，statements，testdata，anthing。要求两个人能完成s和t两个工作，且rank相差2
3 二分图匹配：此题学习建图技巧，两个集和内部一定没有边相连，rank模4小于2和大于等于2的人才能搭配，并且相差正好2，
4 两个人会的不能一样。以上条件删选出两个集合，不能按照anthing来分。
5 */
6 #include <cstdio>
7 #include <iostream>
8 #include <algorithm>
9 #include <cstring>
10 #include <vector>
11 #include <string>
12 #include <cmath>
13 #include <map>
14 using namespace std;
15
16 const int MAXN = 1e3 + 10;
17 const int INF = 0x3f3f3f3f;
18 struct Setter
19 {
20 char name[22], can[11];
21 int rk;
22 bool operator < (const Setter &r) const
23 {
24 return rk < r.rk;
25 }
26 }p[MAXN];
27 int svis[MAXN], vis[MAXN];
28 int mp[MAXN];
29 int x[MAXN], y[MAXN];
30 vector<int> G[MAXN];
31 int n, un, vn;
32
33 bool DFS(int u)
34 {
35 for (int i=0; i<G[u].size (); ++i)
36 {
37 int v = G[u][i];
38 if (!vis[v])
39 {
40 vis[v] = true;
41 if (y[v] == -1 || DFS (y[v]))
42 {
43 y[v] = u; x[u] = v; return true;
44 }
45 }
46 }
47
48 return false;
49 }
50
51 void hungary(void)
52 {
53 sort (p+1, p+1+n); memset (svis, 0, sizeof (svis));
54 un = 0;
55 for (int i=1; i<=n; ++i)
56 {
57 if (svis[i] != 0 || p[i].rk % 4 > 1) continue;
58 svis[i] = -1; mp[++un] = i;
59 for (int j=1; j<=n; ++j)
60 {
61 if (svis[j] == -1 || p[j].rk % 4 <= 1) continue;
62 if (abs (p[i].rk - p[j].rk) != 2) continue;
63 if (p[i].can[0] == p[j].can[0] && p[i].can[0] != 'a') continue;
64 svis[j] = 1; G[un].push_back (j);
65 }
66 }
67
68 int res = 0; memset (x, -1, sizeof (x)); memset (y, -1, sizeof (y));
69 for (int i=1; i<=un; ++i)
70 {
71 memset (vis, false, sizeof (vis));
72 if (DFS (i)) res++;
73 }
74
75 printf ("%d
", res);
76 for (int i=1; i<=un; ++i)
77 {
78 if (x[i] == -1) continue;
79 int id1 = mp[i], id2 = x[i];
80 if (p[id1].can[0] == 't' || p[id2].can[0] == 's') swap (id1, id2);
81 printf ("%s %s
", p[id1].name, p[id2].name);
82 }
83 }
84
85 int main(void) //URAL 1721 Two Sides of the Same Coin
86 {
87 //freopen ("K.in", "r", stdin);
88
89 scanf ("%d", &n);
90 for (int i=1; i<=n; ++i)
91 {
92 scanf ("%s %s %d", p[i].name, p[i].can, &p[i].rk);
93 }
94
95 hungary ();
96
97 return 0;
98 }