HDU-2195-Going Home
http://acm.hdu.edu.cn/showproblem.php?pid=1533
之前用最小費用最大流做的，今天看了最大帶權匹配的KM算法，套用了模板來做這題，將所有的H作爲X集合，所有的m作爲Y集合，構造二分圖求最大帶權匹配
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
const int MAXN=5000;
const int INF=0x7fffffff;
int nx, ny, w[MAXN][MAXN], lx[MAXN], ly[MAXN];
int fx[MAXN], fy[MAXN], matx[MAXN], maty[MAXN];
char map[MAXN][MAXN];
int n1,n2;
struct node
{
int x;
int y;
}h[MAXN],man[MAXN];
//模板
// Name: PerfectMatch by Kuhn_Munkras O(n^3)
// Description: w is the adjacency matrix, nx,ny are the size of x and y,
// lx, ly are the lables of x and y, fx[i], fy[i] is used for marking
// whether the i-th node is visited, matx[x] means x match matx[x],
// maty[y] means y match maty[y], actually, matx[x] is useless,
// all the arrays are start at 1
int path(int u)
{
fx[u] = 1;
for (int v = 1; v <= ny; v++)
if (lx[u] + ly[v] == w[u][v] && fy[v] < 0)
{
fy[v] = 1;
if (maty[v] < 0 || path(maty[v]))
{
matx[u] = v;
maty[v] = u;
return 1;
}
}
return 0;
}
int km()
{
int i,j,k,ret = 0;
memset(ly, 0, sizeof(ly));
for (i = 1; i <= nx; i++)
{
lx[i] = -INF;
for (j = 1; j <= ny; j++)
if (w[i][j] > lx[i]) lx[i] = w[i][j];
}
memset(matx, -1, sizeof(matx));
memset(maty, -1, sizeof(maty));
for (i = 1; i <= nx; i++)
{
memset(fx, -1, sizeof(fx));
memset(fy, -1, sizeof(fy));
if (!path(i))
{
i--;
int p = INF;
for (k = 1; k <= nx; k++)
{
if (fx[k] > 0)
for (j = 1; j <= ny; j++)
if (fy[j] < 0 && lx[k] + ly[j] - w[k][j] < p)
p=lx[k]+ly[j]-w[k][j];
}
for (j = 1; j <= ny; j++)
ly[j] += fy[j]<0 ? 0 : p;
for (j = 1; j <= nx; j++)
lx[j] -= fx[j]<0 ? 0 : p;
}
}
for (i = 1; i <= ny; i++)
ret += w[maty[i]][i];
return ret;
}
void init()
{
int i,j,v;
char c;
nx=ny=0;
for(i=0;i<n1;i++)
{
for(j=0;j<n2;j++)
{
scanf("%c",&c);
if(c=='H')
{
h[++nx].x=i;
h[nx].y=j;
}
else if(c=='m')
{
man[++ny].x=i;
man[ny].y=j;
}
}
getchar();
}
for(i=1;i<=nx;i++)
for(j=1;j<=ny;j++)
w[i][j]=-99999999;
for(i=1;i<=nx;i++)
for(j=1;j<=ny;j++)
w[i][j]=-(abs(h[i].x-man[j].x)+abs(h[i].y-man[j].y));
}
int main()
{
while(scanf("%d %d",&n1,&n2),n1||n2)
{
getchar();
init();
printf("%d\n",-km());
}
return 0;
}