POJ2195 Going Home （最小费最大流||二分图最大权匹配） 2017-02-12 12:14 131人阅读评论(0) 收藏

Going Home

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

————————————————————————————————————

题目的意思是给出一张图，H表示房子，m表示人，人只能上下左右移动一格且花费为1，问所有的人进入房子花费最少是多少？

思路：

方法一：最小费最大流。建图时将每个人和每个房子两两之间建边，流量为1花费为人与房的曼哈顿距离。再加一个源点与每个人建边流量为1花费为0，一个汇点与每个房子建边流量为1花费为0，求源点到汇点的最小花费即可。

方法二：二分图最大权匹配，根据距离关系建立二分图。KM算法解决

方法一：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <string>

#include <queue>

#include <vector>

#include <stack>

#include <set>

#include <map>

using namespace std;

const int MAXN=10000;

const int MAXM=100000;

const int INF=0x3f3f3f3f;

struct Edge{

int to,next,cap,flow,cost;

} edge[MAXM];

int head[MAXN],tol;

int pre[MAXN],dis[MAXN];

bool vis[MAXN];

int N;

struct point{

int x,y;

};

void init(int n)

{

    N=n;

    tol=0;

    memset(head,-1,sizeof head);

}

void addedge(int u,int v,int cap,int cost)

{

    edge[tol].to=v;

    edge[tol].cap=cap;

    edge[tol].flow=0;

    edge[tol].cost=cost;

    edge[tol].next=head[u];

    head[u]=tol++;

    edge[tol].to=u;

    edge[tol].cap=0;

    edge[tol].flow=0;

    edge[tol].cost=-cost;

    edge[tol].next=head[v];

    head[v]=tol++;

}

bool spfa(int s,int t)

{

    queue<int>q;

    for(int i=0;i<N;i++)

    {

        dis[i]=INF;

        vis[i]=false;

        pre[i]=-1;

    }

    dis[s]=0;

    vis[s]=true;

    q.push(s);

    while(!q.empty())

    {

        int u=q.front();

        q.pop();

        vis[u]=false;

        for(int i=head[u];i!=-1;i=edge[i].next)

        {

            int v=edge[i].to;

            if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost)

            {

                dis[v]=dis[u]+edge[i].cost;

                pre[v]=i;

                if(!vis[v])

                {

                    vis[v]=true;

                    q.push(v);

                }

            }

        }

    }

    if(pre[t]==-1)return false;

    return true;

}

int MincostMaxflow(int s,int t)

{

    int flow=0;

    int cost=0;

    while(spfa(s,t))

    {

        int Min=INF;

        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])

        {

            if(Min>edge[i].cap-edge[i].flow)

                Min=edge[i].cap-edge[i].flow;

        }

        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])

        {

            edge[i].flow+=Min;

            edge[i^1].flow-=Min;

            cost+=edge[i].cost*Min;

        }

        flow+=Min;

    }

    return cost;

}

int main()

{

    char mp[105][105];

    int m,n;

    while(~scanf("%d%d",&n,&m)&&(m||n))

    {

        point H[105],P[105];

        int h=0,p=0;

        for(int i=0;i<n;i++)

        {

            scanf("%s",&mp[i]);

            for(int j=0;j<m;j++)

            {

                if(mp[i][j]=='H')

                {

                   H[h].x=i;

                   H[h].y=j;

                   h++;

                }

                else if(mp[i][j]=='m')

                {

                    P[p].x=i;

                    P[p].y=j;

                    p++;

                }

            }

        }

        init(p+h+2);

        for(int i=0;i<h;i++)

            for(int j=0;j<p;j++)

        {

            int c=fabs(H[i].x-P[j].x)+fabs(H[i].y-P[j].y);

             addedge(i+1,h+j+1,1,c);

        }

        for(int i=0;i<h;i++)

        {

            addedge(0,i+1,1,0);

        }

        for(int i=0;i<p;i++)

        {

            addedge(h+1+i,h+p+1,1,0);

        }

        printf("%d\n",MincostMaxflow(0,h+p+1));

    }

}

方法二：

#include <iostream>

#include <cstdio>

#include <string>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <queue>

#include <vector>

#include <set>

#include <stack>

#include <map>

#include <climits>

using namespace std;

#define LL long long

const int INF=0x3f3f3f3f;

const int MAXN = 505;

int g[MAXN][MAXN];

int lx[MAXN],ly[MAXN]; //顶标

int linky[MAXN];

int visx[MAXN],visy[MAXN];

int slack[MAXN];

char mp[MAXN][MAXN];

int nx,ny;

bool find(int x)

{

    visx[x] = true;

    for(int y = 0; y < ny; y++)

    {

        if(visy[y])

            continue;

        int t = lx[x] + ly[y] - g[x][y];

        if(t==0)

        {

            visy[y] = true;

            if(linky[y]==-1 || find(linky[y]))

            {

                linky[y] = x;

                return true;        //找到增广轨

            }

        }

        else if(slack[y] > t)

            slack[y] = t;

    }

    return false;                   //没有找到增广轨（说明顶点x没有对应的匹配，与完备匹配(相等子图的完备匹配)不符）

}

int KM()                //返回最优匹配的值

{

    int i,j;

    memset(linky,-1,sizeof(linky));

    memset(ly,0,sizeof(ly));

    for(i = 0; i < nx; i++)

        for(j = 0,lx[i] = -INF; j < ny; j++)

            lx[i] = max(lx[i],g[i][j]);

    for(int x = 0; x < nx; x++)

    {

        for(i = 0; i < ny; i++)

            slack[i] = INF;

        while(true)

        {

            memset(visx,0,sizeof(visx));

            memset(visy,0,sizeof(visy));

            if(find(x))                     //找到增广轨，退出

                break;

            int d = INF;

            for(i = 0; i < ny; i++)          //没找到，对l做调整(这会增加相等子图的边)，重新找

            {

                if(!visy[i] && d > slack[i])

                    d = slack[i];

            }

            for(i = 0; i < nx; i++)

            {

                if(visx[i])

                    lx[i] -= d;

            }

            for(i = 0; i < ny; i++)

            {

                if(visy[i])

                    ly[i] += d;

                else

                    slack[i] -= d;

            }

        }

    }

    int result = 0;

    for(i = 0; i < ny; i++)

        if(linky[i]>-1)

            result += g[linky[i]][i];

    return result;

}

int main()

{

    int n,m;

    while(~scanf("%d%d",&n,&m)&&(n||m))

    {

        for(int i=0; i<n; i++)

        {

            scanf("%s",mp[i]);

        }

          int cnt=0;

        int CNT=0;

        memset(g,-INF,sizeof g);

        for(int i=0; i<n; i++)

            for(int j=0; j<m; j++)

            {

                if(mp[i][j]=='m')

                {

                    int CNT=0;

                    for(int I=0; I<n; I++)

                        for(int J=0; J<m; J++)

                        {

                            if(mp[I][J]=='H')

                            {

                                g[cnt][CNT++]=-(abs(i-I)+abs(j-J));

                            }

                        }

                    cnt++;

                }

            }

            nx=ny=cnt;

            printf("%d\n",-KM());

    }

        return 0;

    }

巴特西

POJ2195 Going Home （最小费最大流||二分图最大权匹配） 2017-02-12 12:14 131人阅读评论(0) 收藏

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巴特西

POJ2195 Going Home （最小费最大流||二分图最大权匹配） 2017-02-12 12:14 131人阅读 评论(0) 收藏

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POJ2195 Going Home （最小费最大流||二分图最大权匹配） 2017-02-12 12:14 131人阅读评论(0) 收藏