hdu2282：Chocolate（KM匹配）

http://acm.hdu.edu.cn/showproblem.php?pid=2282

Problem Description

Lethe loves eating chocolates very much. In Lethe's birthday, her good friend echo brings N boxes to her, and makes the boxes on the circle. Furthermore, echo tells Lethe that there are many chocolates in the boxes, but the total number of chocolates doesn't exceed N. Also, echo wants Lethe to displace the chocolates in such a way that in each box remains no more than one chocolate. In one move she can shift one chocolate from current box to its neighboring box. (Each box has two neighboring boxes). Can you tell Lethe the minimum number of move to achieve this goal?

Input

There are multi-cases (The total number of cases won't exceed 20). First line is an integer N(1<=N<=500), the total number of boxes. Then N lines follow, each line includes a number, indicates the number of chocolates in the box.

Output

Output the minimum number of move.

Sample Input

10
1
3
3
0
0
2
0
0
0
0

Sample Output

9

题意分析：

有n个盒子，排成环状，里面有若干巧克力，每次可以把一个巧克力移动到相邻的盒子中，要求移动后每个盒子最多只有一个巧克力，求最小的移动次数。

解题思路：

移动肯定是要把盒子中巧克力数大于1的移动到空盒子里，

所以匹配的左边是移动的巧克力，右边是可移动的空盒，权值是移动的最少次数。跑KM匹配即可。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 550
int e[N][N], bookx[N], booky[N], lx[N], ly[N], slack[N], march[N], a[N];
int nx, ny, inf=99999999;
int dfs(int u)
{
	int v, t;
	bookx[u]=1;
	for(v=1; v<=ny; v++)
	{
		if(booky[v]==0)
		{
			t=lx[u]+ly[v]-e[u][v];
			if(t==0)
			{
				booky[v]=1;
				if(march[v]==0 || dfs(march[v]))
				{
					march[v]=u;
					return 1;
				}
			}
			else
				slack[v]=min(slack[v], t);
		}
	}
	return 0;
}
int KM()
{
	int i, j, x, d;
	memset(ly, 0, sizeof(ly));
	memset(march, 0, sizeof(march));
	for(i=1; i<=nx; i++)
	{
		lx[i]=-inf;
		for(j=1; j<=ny; j++)
			lx[i]=max(lx[i], e[i][j]);
	}
	for(x=1; x<=nx; x++)
	{
		for(i=1; i<=ny; i++)
			slack[i]=inf;
		while(1)
		{
			memset(bookx, 0, sizeof(bookx));
			memset(booky, 0, sizeof(booky));
			if(dfs(x))
				break;
			d=inf;
			for(i=1; i<=ny; i++)
				if(!booky[i])
					d=min(d, slack[i]);
			for(i=1; i<=nx; i++)
				if(bookx[i])
					lx[i]-=d;
			for(i=1; i<=ny; i++)
			{
				if(booky[i])
					ly[i]+=d;
				else
					slack[i]-=d;
			}
		}
	}
	int sum=0;
	for(i=1; i<=ny; i++)
		if(march[i])
			sum+=e[march[i]][i];		
	return sum;
}
int main()
{
	int i, j, n, m; 
	while(scanf("%d", &n)!=EOF)
	{
		nx=0;ny=n;
		for(i=1; i<=n; i++)
			scanf("%d", &a[i]);
		for(i=1; i<=n; i++)
		{
			while(a[i]>1)
			{
				nx++;
				a[i]--;
				for(j=1; j<=n; j++)
					if(!a[j])
						e[nx][j]=-min(abs(i-j), n-abs(i-j));
					else
						e[nx][j]=-inf;
			}	
		}
		printf("%d
", -KM());
	}
	return 0;
}