题意：告诉你一些骨牌，然后骨牌的位置与横竖，这样求最多保留多少无覆盖的方格。
这样的话有人用二分匹配，因为两个必定去掉一个，我用的是最小割，因为保证横着和竖着不连通即可。
1 #include <stdio.h> 2 #include <string.h> 3 #include <vector> 4 #include <iostream> 5 #include <queue> 6 #define loop(s,i,n) for(i = s;i < n;i++) 7 using namespace std; 8 const int maxn = 2050; 9 struct edge 10 { 11 int u,v,cap,flow; 12 }; 13 vector<edge>edges; 14 vector<int>g[maxn],G[maxn]; 15 void addedge(int u,int v,int cap,int flow) 16 { 17 //printf("********u %d ****v %d ***** ",u,v); 18 edges.push_back((edge){u,v,cap,flow}); 19 edges.push_back((edge){v,u,0,0}); 20 int m; 21 m = edges.size(); 22 g[u].push_back(m-2); 23 g[v].push_back(m-1); 24 } 25 void init(int n) 26 { 27 int i; 28 for(i = 0;i <= n;i++) 29 { 30 g[i].clear(); 31 } 32 edges.clear(); 33 } 34 int vis[maxn],dis[maxn],cur[maxn]; 35 int bfs(int s,int t,int n) 36 { 37 memset(vis,0,sizeof(vis)); 38 queue<int>q; 39 q.push(s); 40 dis[s] = 0; 41 vis[s] = 1; 42 while(!q.empty()) 43 { 44 int u,v; 45 u = q.front(); 46 q.pop(); 47 for(int i = 0;i < g[u].size();i++) 48 { 49 edge &e = edges[g[u][i]]; 50 v = e.v; 51 if(!vis[v] && e.cap-e.flow > 0) 52 { 53 vis[v] = 1; 54 dis[v] = dis[u] +1; 55 q.push(v); 56 } 57 } 58 } 59 return vis[t]; 60 } 61 int dfs(int u,int a,int t) 62 { 63 if(u == t || a == 0) 64 return a; 65 int v; 66 int flow = 0,f; 67 for(int& i = cur[u]; i < g[u].size();i++) 68 { 69 edge &e = edges[g[u][i]]; 70 v = e.v; 71 if(dis[u]+1 == dis[v] &&(f = dfs(v,min(a,e.cap-e.flow),t))) 72 { 73 e.flow += f; 74 edges[g[u][i]^1].flow -= f; 75 flow += f; 76 a -= f; 77 if(a == 0) 78 break; 79 } 80 } 81 return flow; 82 } 83 int maxflow(int s,int t) 84 { 85 int flow = 0; 86 while(bfs(s,t,t)) 87 { 88 memset(cur,0,sizeof(cur)); 89 flow+=dfs(s,1000050,t); 90 } 91 return flow; 92 } 93 struct node 94 { 95 int x1,y1,x2,y2; 96 }px[2050],py[1050]; 97 int lap(int a,int b) 98 { 99 if(py[a].x1 == px[b].x1 && py[a].y1 == px[b].y1) 100 return 1; 101 if(py[a].x2 == px[b].x1 && py[a].y2 == px[b].y1) 102 return 1; 103 if(py[a].x1 == px[b].x2 && py[a].y1 == px[b].y2) 104 return 1; 105 if(py[a].x2 == px[b].x2 && py[a].y2 == px[b].y2) 106 return 1; 107 return 0; 108 } 109 int main() 110 { 111 int n,m; 112 while(scanf("%d%d",&n,&m)&&(m||n)) 113 { 114 int i,j; 115 int x,y; 116 init(m+n+5); 117 for(i = 0;i < n;i++) 118 { 119 scanf("%d %d",&x,&y); 120 px[i].x1 = x; 121 px[i].y2 = px[i].y1 = y; 122 px[i].x2 = x+1; 123 } 124 for(j = 0;j < m;j++) 125 { 126 scanf("%d %d",&x,&y); 127 py[j].x1 = py[j].x2 = x; 128 py[j].y1 = py[j].y2 = y; 129 py[j].y2++; 130 } 131 int cnt; 132 cnt = 0; 133 loop(0,i,n) 134 addedge(0,i+1,1,0); 135 loop(0,j,m) 136 addedge(n+j+1,m+n+1,1,0); 137 loop(0,i,n) 138 { 139 loop(0,j,m) 140 { 141 if(lap(j,i)) 142 addedge(i+1,j+n+1,1005,0); 143 } 144 } 145 cout<<m+n-maxflow(0,m+n+1)<<endl; 146 } 147 }