带权的二分匹配

#include <queue>
#include <algorithm>
/*********************************************************/
//这些都是KM模板
const int N = 305;//二分图中每一个子图的点的最大数
const int INF = 1<<28;//正无穷

bool xckd
, yckd
;//在一次DFS中，Xi与Yi是否在交错树上
int  n;//点的个案
int edge

;//二维权值信息用矩阵来存储
int xmate
, ymate
;//保存匹配结果
int lx
, ly
;//Xi与Yi和标号，即解说中的A[]和B[]
int slack
;//松弛量
int prev
;//?
queue<int> Q;

bool bfs();//寻找增广路径
void agument(int);
int KMMatch();//KM算法
/*********************************************************/
bool bfs()
{
while(!Q.empty())
{
int p = Q.front(), u = p>>1;
Q.pop();
if(p&1)
{
if(ymate[u] == -1)
{
agument(u);
return true;
}
else
{
xckd[ymate[u]] = true;
Q.push(ymate[u]<<1);
}
}
else
{
for(int i = 0; i < n; i++)
{
if(yckd[i])continue;
else if(lx[u] + ly[i] != edge[u][i])
{
int ex = lx[u] + ly[i] - edge[u][i];
if(slack[i] > ex)
{
slack[i] = ex;
prev[i] = u;
}
}
else
{
yckd[i] = true;
prev[i] = u;
Q.push((i<<1)|1);
}
}
}
}
return false;
}

void agument(int u)
{
while(u != -1)
{
int pv = xmate[prev[u]];
ymate[u] = prev[u];
xmate[prev[u]] = u;
u = pv;
}
}

int KMMatch()
{
int i, j, mn;
memset(ly, 0, sizeof(ly));
for(i = 0; i < n; i++)
{
lx[i] = -INF;
for(j = 0; j < n; j++)
{
lx[i] = lx[i]>edge[i][j]?lx[i]:edge[i][j];
}
}
memset(xmate, -1, sizeof(xmate));
memset(ymate, -1, sizeof(ymate));
bool agu = true;
for(mn = 0; mn < n; mn++)
{
if(agu)
{
memset(xckd, 0, sizeof(xckd));
memset(yckd, 0, sizeof(yckd));
for(i = 0; i < n; i++)
slack[i] = INF;
while(!Q.empty())
Q.pop();
xckd[mn] = true;
Q.push(mn<<1);
}
if(bfs())
{
agu = true;
continue;
}
int ex = INF;
mn--;
agu = false;
for(i = 0; i < n; i++)
if(!yckd[i])
ex = ex<slack[i]?ex:slack[i];
for(i = 0; i < n; i++)
{
if(xckd[i])
lx[i] -= ex;
if(yckd[i])
ly[i] += ex;
slack[i] -= ex;
}
for(i = 0; i < n; i++)
if(!yckd[i] && slack[i] == 0)
{
yckd[i] = true;
Q.push((i<<1)|1);
}
}
int cost = 0;
for(i = 0; i < n; i++)
cost += edge[i][xmate[i]];
return cost;
}