[SCOI2010]连续攻击游戏

题目传送门

这道题是一个二分图匹配问题。如果第$i$件装备的属性值为$j$，将装备$i$连边到属性值$j$（注意装备结点$≠$属性值结点）。

然后从$1$到$10000$跑二分图匹配，判断该点是否能被匹配。直到不能匹配为止。

#include <bits/stdc++.h>

using namespace std;

#define re register
#define rep(i, a, b) for (re int i = a; i <= b; ++i)
#define repd(i, a, b) for (re int i = a; i >= b; --i)
#define maxx(a, b) a = max(a, b);
#define minn(a, b) a = min(a, b);
#define LL long long
#define inf (1 << 30)

inline int read() {
    int w = 0, f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();
    return w * f;
}

const int maxn = 1e6 + 5, maxm = 1e4 + 5;

struct Edge {
    int u, v, pre;
};

struct Graph {
    Edge edges[maxn << 1];
    int G[maxm], n, m;
    void init() {
        memset(G, 0, sizeof(G));
        n = 10000;
        m = 0;
    }
    void Add(int u, int v) {
        edges[++m] = (Edge){u, v, G[u]};
        G[u] = m;
    }
    int vis[maxn], lk[maxn];
    bool dfs(int u, int t) {
        if (vis[u] == t) return 0;
        vis[u] = t;
        for (int i = G[u]; i; i = edges[i].pre) {
            int v = edges[i].v;
            if (!lk[v] || dfs(lk[v], t)) {
                lk[v] = u;
                return true;
            }
        }
        return false;
    }
    int hungarian() {
        rep(i, 1, 10000) {
            if (!dfs(i, i)) // 这是时间戳的一种优化，否则会TLE。
                return i-1;
        }
        return 10000;
    }
} G;

int N;

int main() {
    N = read();
    G.init();
    rep(i, 1, N) {
        int u = read(), v = read();
        G.Add(u, i);
        G.Add(v, i);
    }
    printf("%d", G.hungarian());

    return 0;
}