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 ;
}
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