【HDU 1533】 Going Home （KM）

Going Home

Problem Description

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man. 

Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a '.' means an empty space, an 'H' represents a house on that point, and am 'm' indicates there is a little man on that point. 

You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.

Input

There are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both N and M are between 2 and 100, inclusive. There will be the same number of 'H's and 'm's on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.

Output

For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay. 

Sample Input

2 2 .m H. 5 5 HH..m ..... ..... ..... mm..H 7 8 ...H.... ...H.... ...H.... mmmHmmmm ...H.... ...H.... ...H.... 0 0

Sample Output

2 10 28

Source

Pacific Northwest 2004

 

【题意】

　　

给你一个类似这样的图

...H....

...H....

...H....

mmmHmmmm

...H....

...H....

...H....

问所有H移动到所有m上花费最少的步数

 

【分析】

 

　　这题题解都是费用流，可能不卡费用流。

　　我打的是n^3 KM，不过求的是最小边权的最佳匹配，所以把边权先取反然后再做。

　　记得一开始初始化lx的时候是-INF 不是0，有负边。

 

代码如下：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define Maxn 110
#define Maxm 10010
#define INF 0xfffffff

struct node
{
    int x,y,c,next;
}t[Maxm];int len;
int first[Maxn];

int mymin(int x,int y) {return x<y?x:y;}
int mymax(int x,int y) {return x>y?x:y;}
int myabs(int x) {return x>0?x:-x;}

int hx[Maxn],hy[Maxn],mx[Maxn],my[Maxn];
int fn;

void ins(int x,int y,int c)
{
    t[++len].x=x;t[len].y=y;t[len].c=-c;
    t[len].next=first[x];first[x]=len;
}

int lx[Maxn],ly[Maxn];
bool visx[Maxn],visy[Maxn];
int slack[Maxn],match[Maxn];

char s[Maxn];

bool ffind(int x)
{
    visx[x]=1;
    for(int i=first[x];i;i=t[i].next) if(!visy[t[i].y])
    {
        int y=t[i].y;
        if(lx[x]+ly[y]==t[i].c)
        {
            visy[y]=1;
            if(!match[y]||ffind(match[y]))
            {
                match[y]=x;
                return 1;
            }
        }
        else slack[y]=mymin(slack[y],lx[x]+ly[y]-t[i].c);
    }
    return 0;
}

void solve()
{
    memset(match,0,sizeof(match));
    memset(ly,0,sizeof(ly));
    // memset(lx,0,sizeof(lx));
    for(int i=1;i<=fn;i++)
    {
        lx[i]=-INF;
        for(int j=first[i];j;j=t[j].next) lx[i]=mymax(lx[i],t[j].c);
    }
    
    for(int i=1;i<=fn;i++)
    {
        for(int j=1;j<=fn;j++)
            slack[j]=INF;
        while(1)
        {
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(ffind(i)) break;
            
            int delta=INF;
            for(int j=1;j<=fn;j++) if(!visy[j]) 
              delta=mymin(delta,slack[j]);
            
            if(delta==INF) return ;
            
            for(int j=1;j<=fn;j++)
            {
                if(visx[j]) lx[j]-=delta;
                if(visy[j]) ly[j]+=delta;
                else slack[j]-=delta;
            }
        }
    }
}

int main()
{
    int n,m;
    while(1)
    {
        scanf("%d%d",&n,&m);
        if(n==0&&m==0) break;
        hx[0]=mx[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%s",s);
            for(int j=0;j<m;j++)
            {
                if(s[j]=='H') hx[++hx[0]]=i,hy[hx[0]]=j+1;
                else if(s[j]=='m') mx[++mx[0]]=i,my[mx[0]]=j+1;
            }
        }
        len=0;
        memset(first,0,sizeof(first));
        for(int i=1;i<=hx[0];i++)
         for(int j=1;j<=mx[0];j++)
            ins(i,j,myabs(hx[i]-mx[j])+myabs(hy[i]-my[j]));
        fn=hx[0];
        solve();
        int ans=0;
        for(int i=1;i<=fn;i++) ans+=lx[i]+ly[i];
        printf("%d
",-ans);
    }
    return 0;
}

容易打错的地方是visx 和 visy 的标记。

表示的是是否为增广路上的点，前提当然是他也在相等子图上。

2016-10-27 09:45:40