Going Home hdu 1533 最小费用最大流，费用流好用模板

Going Home

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2416    Accepted Submission(s): 1214

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

 

Recommend

lwg

这个题目真是比较困难，感觉无从下手，这是我第一次做最小费用最大流，想找个

模板实在不容易啊，但是终于找到了好的模板，做出来了，感觉是开心的

以第二组数据为例，是这么建图的

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
int sumFlow;
#define MAXN  1010
#define MAXM  1000200
#define INF  1000000000
struct node
{
    int x,y;
}H[300],M[300];
struct Edge
{
    int u, v, cap, cost;
    int next;
}edge[MAXM<<2];
int NE;
int head[MAXN], dist[MAXN], pp[MAXN];
bool vis[MAXN];
char map[MAXN][MAXN];
void init()
{
    NE = 0;
    memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int cap, int cost)
{
    edge[NE].u = u; edge[NE].v = v; edge[NE].cap = cap; edge[NE].cost = cost;
    edge[NE].next = head[u]; head[u] = NE++;
    edge[NE].u = v; edge[NE].v = u; edge[NE].cap = 0; edge[NE].cost = -cost;
    edge[NE].next = head[v]; head[v] = NE++;
}
bool SPFA(int s, int t, int n)
{
    int i, u, v;
    queue  qu;
    memset(vis,false,sizeof(vis));
    memset(pp,-1,sizeof(pp));
    for(i = 0; i <= n; i++) dist[i] = INF;
    vis[s] = true; dist[s] = 0;
    qu.push(s);
    while(!qu.empty())
    {
        u = qu.front(); qu.pop(); vis[u] = false;
        for(i = head[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].v;
            if(edge[i].cap && dist[v] > dist[u] + edge[i].cost)
            {
                dist[v] = dist[u] + edge[i].cost;
                pp[v] = i;
                if(!vis[v])
                {
                    qu.push(v);
                    vis[v] = true;
                }
            }
        }
    }
    if(dist[t] == INF) return false;
    return true;
}
int MCMF(int s, int t, int n) // minCostMaxFlow
{
    int flow = 0; // 总流量
    int i, minflow, mincost;
    mincost = 0;
    while(SPFA(s, t, n))
    {
        minflow = INF + 1;
        for(i = pp[t]; i != -1; i = pp[edge[i].u])
            if(edge[i].cap < minflow)
                minflow = edge[i].cap;
        flow += minflow;
        for(i = pp[t]; i != -1; i = pp[edge[i].u])
        {
            edge[i].cap -= minflow;
            edge[i^1].cap += minflow;
        }
        mincost += dist[t] * minflow;
    }
    sumFlow = flow; // 最大流
    return mincost;
}
int main()
{
    int n, m;
    int u, v, c;
    while (scanf("%d%d", &n, &m),n|m)
    {
        init();
        int ch=0,cm=0;
        for(int i=0;i