大白书P248有证明,此处贴出两种复杂度的方案，
n^4
大白书P350
n^3
1 #include <algorithm> 2 #include <string.h> 3 #include <iostream> 4 #include <cstdio> 5 using namespace std; 6 /* KM算法 7 * 复杂度O(nx*nx*ny) 8 * 求最大权匹配 9 * 若求最小权匹配，可将权值取相反数，结果取相反数 10 * 点的编号从0开始 11 */ 12 const int N = 310; 13 const int INF = 0x3f3f3f3f; 14 int nx,ny; //两边的点数 15 int W[N][N]; //二分图描述 16 int Left[N],Lx[N],Ly[N]; //y中各点匹配状态， x,y中的点标号 17 int slack[N]; 18 bool S[N],T[N]; 19 bool DFS(int x) { 20 S[x] = true; 21 for(int y = 0; y < ny; y++){ 22 if(T[y]) continue; 23 int tmp = Lx[x] + Ly[y] - W[x][y]; 24 if(tmp == 0){ 25 T[y] = true; 26 if(Left[y] == -1 || DFS(Left[y])){ 27 Left[y] = x; 28 return true; 29 } 30 } 31 else if(slack[y] > tmp) 32 slack[y] = tmp; 33 } 34 return false; 35 } 36 int KM(){ 37 memset(Left, -1, sizeof(Left)); 38 memset(Ly,0, sizeof(Ly)); 39 for(int i = 0;i < nx;i++){ 40 Lx[i] = -INF; 41 for(int j = 0;j < ny;j++) 42 if(W[i][j] > Lx[i]) 43 Lx[i] = W[i][j]; 44 } 45 for(int x = 0;x < nx;x++){ 46 for(int i = 0;i < ny;i++) 47 slack[i] = INF; 48 while(true){ 49 memset(S, false, sizeof(S)); 50 memset(T, false, sizeof(T)); 51 if(DFS(x)) break; 52 int d = INF; 53 for(int i = 0;i < ny;i++) 54 if(!T[i] && d > slack[i]) 55 d = slack[i]; 56 for(int i = 0;i < nx;i++) 57 if(S[i]) 58 Lx[i] -= d; 59 for(int i = 0;i < ny;i++){ 60 if(T[i])Ly[i] += d; 61 else slack[i] -= d; 62 } 63 } 64 } 65 int res = 0; 66 for(int i = 0;i < ny;i++) 67 if(Left[i] != -1) 68 res += W[Left[i]][i]; 69 return res; 70 } 71 //HDU 2255 72 int main() 73 { 74 int n; 75 while(scanf("%d",&n) == 1){ 76 for(int i = 0;i < n;i++) 77 for(int j = 0;j < n;j++) 78 scanf("%d",&W[i][j]); 79 nx = ny = n; 80 printf("%d ",KM()); 81 } 82 return 0; 83 }