最小覆盖数板子题。
每一行看成一个点，每一列也看成一个点，如果(x, y)有障碍，就从点x向y连一条边，这样就构成了一个二分图。然后求最小覆盖数，跑最大匹配。
求最小覆盖数是因为对于这条边连接的连个点，选择任意一个就行，所以求最小覆盖数。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxe = 2e4 + 5; 21 const int maxn = 1e3 + 5; 22 inline ll read() 23 { 24 ll ans = 0; 25 char ch = getchar(), last = ' '; 26 while(!isdigit(ch)) {last = ch; ch = getchar();} 27 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 28 if(last == '-') ans = -ans; 29 return ans; 30 } 31 inline void write(ll x) 32 { 33 if(x < 0) x = -x, putchar('-'); 34 if(x >= 10) write(x / 10); 35 putchar(x % 10 + '0'); 36 } 37 int n, t, k; 38 struct Edge 39 { 40 int nxt, from, to, cap, flow; 41 }e[maxe]; 42 int head[maxn], ecnt = 1; 43 void addEdge(int x, int y) 44 { 45 e[++ecnt] = (Edge){head[x], x, y, 1, 0}; 46 head[x] = ecnt; 47 e[++ecnt] = (Edge){head[y], y, x, 0, 0}; 48 head[y] = ecnt; 49 } 50 51 int dis[maxn]; 52 bool bfs() 53 { 54 Mem(dis, 0); dis[0] = 1; 55 queue<int> q; q.push(0); 56 while(!q.empty()) 57 { 58 int now = q.front(); q.pop(); 59 for(int i = head[now]; i; i = e[i].nxt) 60 if(!dis[e[i].to] && e[i].cap > e[i].flow) 61 { 62 dis[e[i].to] = dis[now] + 1; 63 q.push(e[i].to); 64 } 65 } 66 return dis[t]; 67 } 68 int cur[maxn]; 69 int dfs(int now, int res) 70 { 71 if(now == t || res == 0) return res; 72 int flow = 0, f; 73 if(!cur[now]) cur[now] = head[now]; 74 for(int &i = cur[now]; i; i = e[i].nxt) 75 { 76 if(dis[e[i].to] == dis[now] + 1 && (f = dfs(e[i].to, min(res, e[i].cap - e[i].flow))) > 0) 77 { 78 e[i].flow += f; e[i ^ 1].flow -= f; 79 flow += f; res -= f; 80 if(res == 0) break; 81 } 82 } 83 return flow; 84 } 85 86 int maxflow() 87 { 88 int flow = 0; 89 while(bfs()) 90 { 91 Mem(cur, 0); 92 flow += dfs(0, INF); 93 } 94 return flow; 95 } 96 97 int main() 98 { 99 n = read(); k = read(); t = n + n + 1; 100 for(int i = 1; i <= n; ++i) addEdge(0, i), addEdge(i + n, t); 101 for(int i = 1; i <= k; ++i) 102 { 103 int x = read(), y = read(); 104 addEdge(x, y + n); 105 } 106 write(maxflow()), enter; 107 return 0; 108 }