POJ 2195 Going Home(费用流)

题意给定一个N*M的地图,地图上有若干个man和house,且man与house的数量一致。man每移动一格需花费$1,一间house只能入住一个man。现在要求所有的man都入住house,求最小费用。

思路:明显的最小费用最大流,源点为0,每个人为一个点连源点,容量为1,边权(费用)为0,每个人连每一间房子,容量为1,边权为两者坐标之差的和,然后房子和汇点连边,容量为1,边权为0即可,非常简单的建图和思路,然后跑一遍最小费用最大流就可以了。

Trick:这么简单的水题我居然超时了快十次,一开始以为是我用string数组来处理输入超时,改用了char数组,还是超时,就不懂了....甚至怀疑模板敲错了..又敲了一次..还是TLE..最后折腾了一晚上,最后我把初始化时候将按照输入的n初始化改成了统计的人的数目来初始化,80多MS就过了...gg...题目不是给了n最多不是也才100么..表示懵逼...


#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 220
#define LL long long
int cas=1,T;
const int INF = 1e9;
struct Edge
{
	int from,to,cap,flow,cost;
	Edge(){}
	Edge(int u,int v,int c,int f,int co):from(u),to(v),cap(c),flow(f),cost(co){}
};
int n,m;
int num1,num2;
struct MCMF
{
	int s,t;
	vector<Edge>edges;
	vector<int>G[maxn];
	int inq[maxn];         //是否在队列中
	int d[maxn];
	int p[maxn];           //上一条弧
	int a[maxn];           //可改进量

	void init()
	{
		for (int i = 0;i<=num1*2+1;i++)
			G[i].clear();
		edges.clear();
	}

	void AddEdge(int from,int to,int cap,int cost)
	{
		edges.push_back(Edge(from,to,cap,0,cost));
		edges.push_back(Edge(to,from,0,0,-cost));
		int mm = edges.size();
		G[from].push_back(mm-2);
		G[to].push_back(mm-1);
	}

	bool BellmanFord(int &flow,int &cost)
	{
		for (int i = 0;i<=2*num1+1;i++)
			d[i]=INF;
		memset(inq,0,sizeof(inq));
		d[s]=0;
		inq[s]=1;
		p[s]=0;
		a[s]=INF;
		queue<int>q;
		q.push(s);
		while (!q.empty())
		{
			int u = q.front();q.pop();
			inq[u]=0;
			for (int i = 0;i<G[u].size();i++)
			{
				Edge &e = edges[G[u][i]];
				if (e.cap > e.flow && d[e.to]>d[u]+e.cost)
				{
					d[e.to]=d[u]+e.cost;
					p[e.to]=G[u][i];
					a[e.to]=min(a[u],e.cap-e.flow);
					if (!inq[e.to])
					{
						q.push(e.to);
						inq[e.to]=1;
					}
				}
			}
		}
		if (d[t]==INF)
			return false;        //s-t不连通,失败退出
		flow+=a[t];
		cost+=d[t]*a[t];
		int u = t;
		while (u!=s)
		{
			edges[p[u]].flow+=a[t];
			edges[p[u]^1].flow-=a[t];
			u=edges[p[u]].from;
		}
		return true;
	}

	int Mincost(int s,int t)
	{
		this->s=s;
		this->t=t;
		int flow = 0;
		int cost = 0;
		while (BellmanFord(flow,cost));
		return cost;
	}
}mc;

string mapp[120];
//char mapp[120][120];
struct peo
{
	int x,y;
}p[120];
struct hou
{
	int x,y;
}h[120];
int main()
{
	while (scanf("%d%d",&n,&m)==2 && n)
	{
		int house = 0;
		int peo = 0;
		int hpos = 0;
		int peopos = 0;
         for (int i = 0;i<n;i++)
			 cin >> mapp[i];
         for (int i = 0;i<n;i++)
			 for (int j = 0;j<m;j++)
			 {
				 if (mapp[i][j] == 'm')
                 {
                     peo++;
                     p[peopos].x=i;
					 p[peopos++].y=j;
				 }
				 if (mapp[i][j] == 'H')
				 {
					 house++;
					 h[hpos].x=i;
					 h[hpos++].y=j;
				 }
			 }
         num1 = peo;
		 mc.init();
		 for (int i = 0;i<peo;i++)
		 {
			 mc.AddEdge(0,i+1,1,0);
			 mc.AddEdge(i+peo+1,peo*2+1,1,0);
			 for (int j = 0;j<house;j++)
			 {
				 mc.AddEdge(i+1,peo+1+j,1,(abs(p[i].x-h[j].x)+abs(p[i].y-h[j].y)));
			 }
		 }
		 printf("%d\n",mc.Mincost(0,2*peo+1));
	}
	return 0;
}


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


你可能感兴趣的:(POJ 2195 Going Home(费用流))