POJ 2195-Going Home（KM演算法/最小費用最大流演算法）

kewlgrl發表於2016-08-30
Go演算法

Going Home
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 21730   Accepted: 10990

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

題目意思：

N行M列，每人可分得一個房子，每人每走一步花費為1，求每人各回各家的最小花費。

解題思路：

這題有兩種思路可破：KM演算法和最小費用最大流演算法。

（一）

套Kuhn-Munkras演算法模板，建二分圖求解。

下面詳細講講建圖的過程：

①掃描讀入的圖，分別標記出人M和房屋H的位置座標；

②每個人與每間房屋一一對應，比如測試用例

5 5
HH..m
.....
.....
.....
mm..H
M的座標是A(1,5) B(5,1) C(5,2)；H的座標是X(1,1) Y(1,2) Z(5,5)；

那麼分別求出AX AY AZ BX BY BZ CX CY CZ路徑上對應花費的權值就是二分圖中對應邊的邊權。
因為是求最小權，所以建圖時邊的權值取相反數，最後所得的ans取相反數。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<malloc.h>
using namespace std;
typedef long long ll;
#define maxn 300
#define INF 0x3f3f3f3f
int nx,ny;//兩邊的點數
int g[maxn][maxn];//二分圖描述
int linker[maxn],lx[maxn],ly[maxn];//y中各點匹配狀態，x,y中的點標號
int slack[maxn];
bool visx[maxn],visy[maxn];
int ax[maxn],ay[maxn],bx[maxn],by[maxn];
char Map[maxn][maxn];
bool DFS(int x)
{
    visx[x] = true;
    for(int y = 1; y <= ny; y++)
    {
        if(visy[y])continue;
        int tmp = lx[x] + ly[y] - g[x][y];
        if(tmp == 0)
        {
            visy[y] = true;
            if(linker[y] == -1 || DFS(linker[y]))
            {
                linker[y] = x;
                return true;
            }
        }
        else if(slack[y] > tmp)
            slack[y] = tmp;
    }
    return false;
}
int KM()
{
    memset(linker,-1,sizeof(linker));
    memset(lx,0,sizeof(lx));
    memset(ly,0,sizeof(ly));
    for(int i = 0; i < nx; i++)
    {
        lx[i] = -INF;
        for(int j = 1; j <= ny; j++)
            if(g[i][j] > lx[i])
                lx[i] = g[i][j];
    }
    for(int x = 1; x <= nx; x++)
    {
        for(int i = 1; i <= ny; i++)
            slack[i] = INF;
        while(true)
        {
            memset(visx,false,sizeof(visx));
            memset(visy,false,sizeof(visy));
            if(DFS(x))break;
            int d = INF;
            for(int i = 1; i <= ny; i++)
                if(!visy[i] && d > slack[i])
                    d = slack[i];
            for(int i = 1; i <= nx; i++)
                if(visx[i])
                    lx[i] -= d;
            for(int i = 1; i <= ny; i++)
            {
                if(visy[i])ly[i] += d;
                else slack[i] -= d;
            }
        }
    }
    int res = 0;
    for(int i = 1; i <= nx; i++)
        if(linker[i] != -1)
            res += g[linker[i]][i];
    return -res;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)&&n!=0&&m!=0)
    {
        /*for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++)
                scanf("%d",&g[i][j]);
        nx = ny = n;*/
        memset(g,0,sizeof g);
        nx=ny=0;//二分圖頂點的個數
        for(int i=1; i<=n; i++)
            scanf("%s",Map[i]+1);//讀入原圖
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
            {
                if(Map[i][j]=='m')
                {
                    nx++;
                    ax[nx]=i;
                    ay[nx]=j;
                }
                else if(Map[i][j]=='H')
                {
                    ny++;
                    bx[ny]=i;
                    by[ny]=j;
                }
            }
        for(int i=1; i<=nx; i++)
            for(int j=1; j<=ny; j++)
                g[i][j]=-(abs(ax[i]-bx[j])+abs(ay[i]-by[j]));//建圖
        printf("%d\n",KM());
    }
    return 0;
}
/**
2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0
**/

（二）

套最小費用最大流演算法模板，加超級源點和超級匯點網路流求解。

注意模板使用的時候，minCostMaxflow()的返回值。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<queue>
#include<malloc.h>
using namespace std;
typedef long long ll;
#define maxn 300

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;//節點總個數，節點編號從0~N-1
void init(int n)
{
    N = n;
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)//加邊
{
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].cost = -cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
bool spfa(int s,int t)//找增廣路
{
    queue<int>q;
    for(int i = 0; i <= t+2; 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;
    else return true;
}
//int minCostMaxflow(int s,int t,int &cost)
int minCostMaxflow(int s,int t)//最小費用最大流
{
    int flow = 0;
    int ans=0;
    //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;
        }
        ans+=dis[t];
        flow += Min;
    }
    return ans;
}

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)&&n!=0&&m!=0)
    {
        int nx,ny,ax[maxn],ay[maxn],bx[maxn],by[maxn];
        char Map[maxn][maxn];
        nx=ny=0;//二分圖頂點的個數
        for(int i=1; i<=n; i++)
            scanf("%s",Map[i]+1);//讀入原圖
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
            {
                if(Map[i][j]=='m')
                {
                    nx++;
                    ax[nx]=i;
                    ay[nx]=j;
                }
                else if(Map[i][j]=='H')
                {
                    ny++;
                    bx[ny]=i;
                    by[ny]=j;
                }
            }
        int s=0,t=nx+ny+1;//源點和匯點
        init(t+1);//初始化
        for(int i = 1; i <= nx; i++)
            addedge(s,i,1,0);//源點向各個點加邊
        for(int i = nx+1; i <= nx+ny; i++)
            addedge(i,t,1,0);//匯點向各個點加邊
        for(int i=1; i<=nx; i++)
            for(int j=1; j<=ny; j++)
            {
                int va=(abs(ax[i]-bx[j])+abs(ay[i]-by[j]));
                //cout<<i<<"->"<<nx+j<<"="<<va<<endl;
                addedge(i,nx+j,1,va);//建圖
            }
        printf("%d\n",minCostMaxflow(s,t));
    }
    return 0;
}
/**
2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0
**/


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