题目链接：http://poj.org/problem?id=3041
题意：N*N中选最少行或者列，可以覆盖所有的点。
最小点覆盖问题，等于最大匹配。
建图：考虑行，选中某一行，那么行上的点对应的列则不需要被选。根据这一条，建图就行。
1 #include <algorithm> 2 #include <iostream> 3 #include <iomanip> 4 #include <cstring> 5 #include <climits> 6 #include <complex> 7 #include <cassert> 8 #include <cstdio> 9 #include <bitset> 10 #include <vector> 11 #include <deque> 12 #include <queue> 13 #include <stack> 14 #include <ctime> 15 #include <set> 16 #include <map> 17 #include <cmath> 18 using namespace std; 19 20 typedef pair<int, int> pii; 21 const int maxn = 1050; 22 const int inf = 0x3f3f3f3f; 23 int n, k; 24 int G[maxn][maxn]; 25 int nx,ny; 26 int linker[maxn],lx[maxn],ly[maxn]; 27 int slack[maxn]; 28 bool visx[maxn],visy[maxn]; 29 30 bool dfs(int x) { 31 visx[x] = true; 32 for(int y = 0; y < ny; y++) { 33 if(visy[y])continue; 34 int tmp = lx[x] + ly[y] - G[x][y]; 35 if(tmp == 0) { 36 visy[y] = true; 37 if(linker[y] == -1 || dfs(linker[y])) { 38 linker[y] = x; 39 return true; 40 } 41 } 42 else if(slack[y] > tmp) 43 slack[y] = tmp; 44 } 45 return false; 46 } 47 48 int km() { 49 memset(linker,-1,sizeof(linker)); 50 memset(ly,0,sizeof(ly)); 51 for(int i = 0;i < nx;i++) { 52 lx[i] = -inf; 53 for(int j = 0;j < ny;j++) 54 if(G[i][j] > lx[i]) 55 lx[i] = G[i][j]; 56 } 57 for(int x = 0;x < nx;x++) { 58 for(int i = 0;i < ny;i++) 59 slack[i] = inf; 60 while(true) { 61 memset(visx,false,sizeof(visx)); 62 memset(visy,false,sizeof(visy)); 63 if(dfs(x))break; 64 int d = inf; 65 for(int i = 0;i < ny;i++) 66 if(!visy[i] && d > slack[i]) 67 d = slack[i]; 68 for(int i = 0;i < nx;i++) 69 if(visx[i]) 70 lx[i] -= d; 71 for(int i = 0;i < ny;i++) { 72 if(visy[i])ly[i] += d; 73 else slack[i] -= d; 74 } 75 } 76 } 77 int res = 0; 78 for(int i = 0;i < ny;i++) 79 if(linker[i] != -1) 80 res += G[linker[i]][i]; 81 return res; 82 } 83 84 int main() { 85 // freopen("in", "r", stdin); 86 int x, y; 87 while(~scanf("%d%d",&n,&k)) { 88 memset(G, 0, sizeof(G)); 89 for(int i = 1; i <= k; i++) { 90 scanf("%d%d",&x,&y); 91 G[x-1][y-1]++; 92 } 93 nx = n, ny = n; 94 printf("%d ", km()); 95 } 96 return 0; 97 }