題目傳送門

這道題是一個二分圖匹配問題。如果第$i$件裝備的屬性值為$j$，將裝備$i$連邊到屬性值$j$（註意裝備結點$≠$屬性值結點）。

然後從$1$到$10000$跑二分圖匹配，判斷該點是否能被匹配。直到不能匹配為止。

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define re register
 6 #define rep(i, a, b) for (re int i = a; i <= b; ++i)
 7 #define repd(i, a, b) for (re int i = a; i >= b; --i)
 8
 
 #define maxx(a, b) a = max(a, b);
 9 #define minn(a, b) a = min(a, b);
10 #define LL long long
11 #define inf (1 << 30)
12 
13 inline int read() {
14     int w = 0, f = 1; char c = getchar();
15     while (!isdigit(c)) f = c == ‘-‘ ? -1 : f, c = getchar();
16     while (isdigit(c)) w = (w << 3
 
) + (w << 1) + (c ^ ‘0‘), c = getchar();
17     return w * f;
18 }
19 
20 const int maxn = 1e6 + 5, maxm = 1e4 + 5;
21 
22 struct Edge {
23     int u, v, pre;
24 };
25 
26 struct Graph {
27     Edge edges[maxn << 1];
28     int G[maxm], n, m;
29     void init() {
30         memset(G, 0
 
, sizeof(G));
31         n = 10000;
32         m = 0;
33     }
34     void Add(int u, int v) {
35         edges[++m] = (Edge){u, v, G[u]};
36         G[u] = m;
37     }
38     int vis[maxn], lk[maxn];
39     bool dfs(int u, int t) {
40         if (vis[u] == t) return 0;
41         vis[u] = t;
42         for (int i = G[u]; i; i = edges[i].pre) {
43             int v = edges[i].v;
44             if (!lk[v] || dfs(lk[v], t)) {
45                 lk[v] = u;
46                 return true;
47             }
48         }
49         return false;
50     }
51     int hungarian() {
52         rep(i, 1, 10000) {
53             if (!dfs(i, i)) // 這是時間戳的一種優化，否則會TLE。
54                 return i-1;
55         }
56         return 10000;
57     }
58 } G;
59 
60 int N;
61 
62 int main() {
63     N = read();
64     G.init();
65     rep(i, 1, N) {
66         int u = read(), v = read();
67         G.Add(u, i);
68         G.Add(v, i);
69     }
70     printf("%d", G.hungarian());
71 
72     return 0;
73 }

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