我们将每一个属性和物品连边，然后枚举从小到大属性跑匈牙利，直到找不到连边
#include#include #include #define N 1000001#define M 2000001using namespace std;int n, cnt;int head[N], to[M], nex[M], belong[N];bool vis[N];inline int read(){ int x = 0, f = 1; char ch = getchar(); for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1; for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0'; return x * f;}inline void add(int x, int y){ to[cnt] = y; nex[cnt] = head[x]; head[x] = cnt++;}inline bool dfs(int u){ int i, v; for(i = head[u]; ~i; i = nex[i]) { v = to[i]; if(!vis[v]) { vis[v] = 1; if(!belong[v] || dfs(belong[v])) { belong[v] = u; return 1; } } } return 0;}inline int solve(){ int i, ans = 0; for(i = 1; i <= 10000; i++) { memset(vis, 0, sizeof(vis)); if(dfs(i)) ans++; else return ans; } return ans;}int main(){ int i, x, y; n = read(); memset(head, -1, sizeof(head)); for(i = 1; i <= n; i++) { x = read(); y = read(); add(x, i); add(y, i); } printf("%d\n", solve()); return 0;}