题目是跟 zoj1516是一样的，但多了匹配后的输出
详解zoj1516可见http://www.cnblogs.com/CSU3901130321/p/4228057.html
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 using namespace std; 5 6 const int maxn = 55; 7 int n , m , k; 8 int g[maxn][maxn] , cx[maxn] , cy[maxn] , visy[maxn] , nx , ny; 9 bool col[105][105];//判断是否为鱼塘 10 11 struct Point{ 12 int x , y; 13 Point(int x=0 , int y=0):x(x),y(y){} 14 }; 15 16 Point px[maxn] , py[maxn]; 17 18 bool ok(Point p1 , Point p2) 19 { 20 if((p1.x == p2.x && abs(p1.y - p2.y) == 1) || (p1.y == p2.y && abs(p1.x - p2.x) == 1)) 21 return true; 22 return false; 23 } 24 25 void build_graph() 26 { 27 memset(g , 0 , sizeof(g)); 28 for(int i=1 ; i<=nx ; i++){ 29 for(int j=1 ; j<=ny ; j++){ 30 if(ok(px[i] , py[j])) 31 g[i][j] = 1; 32 } 33 } 34 } 35 36 int dfs(int u) 37 { 38 for(int v = 1 ; v<=ny ; v++) 39 if(g[u][v] && !visy[v]){ 40 visy[v] = 1; 41 if(cy[v] == -1 || dfs(cy[v])){ 42 cx[u] = v; 43 cy[v] = u; 44 return 1; 45 } 46 } 47 return 0; 48 } 49 50 int MaxMatch() 51 { 52 memset(cx , -1 , sizeof(cx)); 53 memset(cy , -1 , sizeof(cy)); 54 int ans = 0; 55 for(int i=1 ; i<=nx ; i++){ 56 if(cx[i] == -1){ 57 memset(visy , 0 , sizeof(visy)); 58 ans += dfs(i); 59 } 60 } 61 return ans; 62 } 63 64 int main() 65 { 66 // freopen("a.in" , "r" ,stdin); 67 while(scanf("%d%d" , &n , &m) , n) 68 { 69 scanf("%d" , &k); 70 int x , y; 71 memset(col , 0 , sizeof(col)); 72 while(k--){ 73 scanf("%d%d" , &x , &y); 74 col[x][y] = 1; 75 } 76 nx = 0 , ny = 0; 77 for(int i=1 ; i<=n ; i++) 78 for(int j=1 ; j<=m ; j++) 79 if(!col[i][j]){ 80 if((i+j)&1) py[++ny] = Point(i,j); 81 else px[++nx] = Point(i,j); 82 } 83 //构造二分图 84 build_graph(); 85 printf("%d " , MaxMatch()); 86 for(int i=1 ; i<=nx ; i++){ 87 if(cx[i] != -1) 88 printf("(%d,%d)--(%d,%d) ",px[i].x , px[i].y , py[cx[i]].x , py[cx[i]].y); 89 } 90 } 91 return 0; 92 }