KM算法模板

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

 #include <algorithm>
 #include <string.h>
 #include <iostream>
 #include <cstdio>
 using namespace std;
 /* KM算法
 * 复杂度O(nx*nx*ny)
 * 求最大权匹配
 * 若求最小权匹配，可将权值取相反数，结果取相反数
 * 点的编号从0开始
 */
 const int N = ;
 const int INF = 0x3f3f3f3f;
 int nx,ny; //两边的点数
 int W[N][N]; //二分图描述
 int Left[N],Lx[N],Ly[N]; //y中各点匹配状态， x,y中的点标号
 int slack[N];
 bool S[N],T[N];
 bool DFS(int x) {
     S[x] = true;
     for(int y = ; y < ny; y++){
         if(T[y]) continue;
         int tmp = Lx[x] + Ly[y] - W[x][y];
         if(tmp == ){
             T[y] = true;
             if(Left[y] == - || DFS(Left[y])){
                 Left[y] = x;
                 return true;
             }
         }
         else if(slack[y] > tmp)
         slack[y] = tmp;
     }
     return false;
 }
 int KM(){
     memset(Left, -, sizeof(Left));
     memset(Ly,, sizeof(Ly));
     for(int i = ;i < nx;i++){
     Lx[i] = -INF;
     for(int j = ;j < ny;j++)
         if(W[i][j] > Lx[i])
             Lx[i] = W[i][j];
     }
     for(int x = ;x < nx;x++){
         for(int i = ;i < ny;i++)
             slack[i] = INF;
         while(true){
             memset(S, false, sizeof(S));
             memset(T, false, sizeof(T));
             if(DFS(x)) break;
             int d = INF;
             for(int i = ;i < ny;i++)
             if(!T[i] && d > slack[i])
             d = slack[i];
             for(int i = ;i < nx;i++)
                 if(S[i])
             Lx[i] -= d;
             for(int i = ;i < ny;i++){
                 if(T[i])Ly[i] += d;
                 else slack[i] -= d;
             }
         }
     }
     int res = ;
     for(int i = ;i < ny;i++)
         if(Left[i] != -)
             res += W[Left[i]][i];
     return res;
 }
 //HDU 2255
 int main()
 {
     int n;
     while(scanf("%d",&n) == ){
         for(int i = ;i < n;i++)
             for(int j = ;j < n;j++)
            scanf("%d",&W[i][j]);
         nx = ny = n;
         printf("%d\n",KM());
     }
     return ;
 }