网络流 ( 最小费用最大流 )——Going Home ( HDU 1533 )

• 题目链接：
  http://acm.hdu.edu.cn/showproblem.php?pid=1533

• 分析：
  给出一个网格图，H代表房子，m代表人，两者数量相等，人每移动一格花费1$，求所有人走到房子里所需要的最小花费。利用最小费用最大流来做。（需要注意的是abs函数可能需要自己写）

• 题解：

  1. 建图：
     ①从S点到每个人建边，容量为1，费用为0；
     ②从每个人到每个房子建边，容量为1，费用为曼哈顿距离；
     ③从每个房子到T建边，容量为1，费用为0。

int NE,head[maxn];
struct Edge
{
    int u,v, flow, cost; //flow为流量 cost为费用
    int  next;
} edge[maxm];

void add(int u,int v,int cap,int cost)
{
        edge[NE].u=u;edge[NE].v=v;edge[NE].flow=cap;edge[NE].cost=cost;
        edge[NE].next=head[u];head[u]=NE++;
        edge[NE].u=v;edge[NE].v=u;edge[NE].flow=0;edge[NE].cost=-cost;
        edge[NE].next=head[v];head[v]=NE++;
}

void init()
{
    memset(head,-1,sizeof(head));
    NE=0;
    ans = 0;
}

2.SPFA算法遍历图找最大流最小费用：

bool vis[maxn];
int dist[maxn], pre[maxn];

int s, t, aug;
int n; //记录的是总的边数
bool spfa()
{
    int k,p,V;
    queue<int> q;
     memset(pre, -1,sizeof(pre));
    memset(vis, 0, sizeof (vis));
    for (int i = 0; i <=n; i++)
        dist[i] = INF;
    q.push(s);
    vis[s] = 1;
    dist[s] = 0;
    while (!q.empty())
    {
        k = q.front();
        q.pop();
        vis[k] = 0;
        for (p=head[k]; p!=-1; p=edge[p].next)
        {
            V = edge[p].v;
            if (edge[p].flow && (edge[p].cost + dist[k] < dist[V]))
            {
                dist[V] = edge[p].cost + dist[k];
                pre[V] = p;
                if (!vis[V])
                {
                    q.push(V);
                    vis[V] = 1;
                }
            }
        }
    }
    if (dist[t] == INF) return false;
    aug = INF+1;
    for (p=pre[t]; p!=-1; p=pre[edge[p].u])
    {
        aug = min(aug, edge[p].flow);
    }
    for (p=pre[t]; p!=-1; p=pre[edge[p].u])
    {
        edge[p].flow -= aug;
        edge[p^1].flow += aug;
    }
    ans += dist[t] * aug;
    return true;
}
 //最后计算时这样写
 while( spfa());//不断找最短路
printf("%d\n", ans);//找不到就输出

• AC 代码：

#include 
#include 
#include 
#include 
#include 
#define Clear(a) memset(a,0,sizeof(a))
using namespace std;

const int maxn = 101000;
const int maxm = 1000200;;
const int INF = 1000000000;

int absI(int x)
{
    if(x<0)
        x = -x;
    return x;
}

struct Edge
{
    int u,v, flow, cost; //flow为流量 cost为费用
    int  next;
} edge[maxm];

int ans, NE;
int head[maxn];

void add(int u,int v,int cap,int cost)
    {
        edge[NE].u=u;edge[NE].v=v;edge[NE].flow=cap;edge[NE].cost=cost;
        edge[NE].next=head[u];head[u]=NE++;
        edge[NE].u=v;edge[NE].v=u;edge[NE].flow=0;edge[NE].cost=-cost;
        edge[NE].next=head[v];head[v]=NE++;
    }

void init()
{
    memset(head,-1,sizeof(head));
        NE=0;
        ans = 0;
}


bool vis[maxn];
int dist[maxn], pre[maxn];

int s, t, aug;
int n;
bool spfa()
{
    int k,p,V;
    queue<int> q;
     memset(pre, -1,sizeof(pre));
    memset(vis, 0, sizeof (vis));
    for (int i = 0; i <=n; i++)
        dist[i] = INF;
    q.push(s);
    vis[s] = 1;
    dist[s] = 0;
    while (!q.empty())
    {
        k = q.front();
        q.pop();
        vis[k] = 0;
        for (p=head[k]; p!=-1; p=edge[p].next)
        {
            V = edge[p].v;
            if (edge[p].flow && (edge[p].cost + dist[k] < dist[V]))
            {
                dist[V] = edge[p].cost + dist[k];
                //cout << "DIST:" << dist[V] << endl;
                pre[V] = p;
                if (!vis[V])
                {
                    q.push(V);
                    vis[V] = 1;
                }
            }
        }
    }
    //cout << dist[t] << endl;
    if (dist[t] == INF) return false;
    aug = INF+1;
    for (p=pre[t]; p!=-1; p=pre[edge[p].u])
    {
        aug = min(aug, edge[p].flow);
        //cout << aug << endl;
    }
    for (p=pre[t]; p!=-1; p=pre[edge[p].u])
    {
        edge[p].flow -= aug;
        edge[p^1].flow += aug;
    }
    ans += dist[t] * aug;
    //cout << ans << endl;
    return true;
}
struct node
{
    int x, y;
};
char Map[110][110];
node man[110];
node House[110];
int main()
{
    int nn,mm;
    while(scanf("%d%d", &nn, &mm) && (nn+mm) )
    {
        getchar();
        init();
        int mans = 0;
        int houses = 0;
        for(int i=1;i<=nn;i++)
        {
            for(int j=1;j<=mm;j++)
            {
                Map[i][j] = getchar();
                if(Map[i][j] == 'm')
                {
                    mans++;
                    man[mans].x = i;
                    man[mans].y = j;
                }
                if(Map[i][j] == 'H')
                {
                    houses++;
                    House[houses].x = i;
                    House[houses].y = j;
                }
            }
            getchar();
        }

        s = 0;
        t = mans+houses+1;
        //cout << t << endl;
        n = 0;
        for(int i=1; i<=mans; i++)
        {
            add(s, i, 1, 0);
            n++;
        }
        for(int i=1; i<=mans; i++)
        {
            for(int j=1; j<=houses; j++)
            {
                int distant = absI( House[j].x - man[i].x)+absI(House[j].y - man[i].y);
                //cout <<"KK "<
                add(i, mans+j, 1, distant);
                n++;
            }
        }
        for(int j=1; j<=houses; j++)
        {
            add(mans+j, t, 1, 0);
            n++;
        }
        ans = 0;
        while(spfa() );
        printf("%d\n", ans);
    }
    return 0;
}