UVA-1045 - The Great Wall Game(二分图最佳匹配)

题意:在一个n*n的棋盘上有n个棋子，要求通过移动棋子使棋子的排布满足以下情况之一：呈横行排列；呈纵行排列；呈对角线排列(有两条)。

棋子移动一个单元格的费用为1，总费用为所有棋子的移动费用之和。求最小费用。

分析:因为这题的数据很小,故可以枚举每种情况.对于每种情况,我们可以用二分匹配的方法算出最小费用(对于每个棋子,连接n条边到n个目标位置,权值设为负,这样最佳二分匹配求的最大值就是答案的最小值了);

// File Name: 1045.cpp
// Author: Zlbing
// Created Time: 2013/4/20 15:58:22

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 1000000000
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
const int MAXN=20;
int X[MAXN],Y[MAXN];
int Left[MAXN];
int w[MAXN][MAXN];
int Lx[MAXN],Ly[MAXN];
bool S[MAXN],T[MAXN];
int N;
bool match(int i)
{S[i]=true;for(int j=1;j<=N;j++)if(Lx[i]+Ly[j]==w[i][j]&&!T[j]){T[j]=true;if(Left[j]==0||match(Left[j])){Left[j]=i;return true;}}return false;
}
void update(){int a=INF;for(int i=1;i<=N;i++)if(S[i])for(int j=1;j<=N;j++)if(!T[j])a=min(a,Lx[i]+Ly[j]-w[i][j]);for(int i=1;i<=N;i++){if(S[i])Lx[i]-=a;if(T[i])Ly[i]+=a;}
}
void KM()
{for(int i=1;i<=N;i++){Left[i]=Lx[i]=Ly[i]=0;for(int j=1;j<=N;j++){Lx[i]=max(Lx[i],w[i][j]);}}for(int i=1;i<=N;i++){for(;;){CL(S,0);CL(T,0);if(match(i))break;else update();}}
}int main()
{int cas=0;    while(~scanf("%d",&N)){if(N==0)break;REP(i,1,N){int a,b;scanf("%d%d",&a,&b);X[i]=a;Y[i]=b;}int ans=INF;REP(i,1,N){REP(k,1,N)REP(j,1,N){int dist=abs(X[k]-i)+abs(Y[k]-j);w[k][j]=-dist;}KM();int minn=0;REP(i,1,N){minn+=-w[Left[i]][i];}ans=min(ans,minn);}REP(i,1,N) {REP(k,1,N)REP(j,1,N){int dist=abs(X[k]-j)+abs(Y[k]-i);w[k][j]=-dist;}KM();int minn=0;REP(i,1,N)minn+=-w[Left[i]][i];ans=min(ans,minn);}REP(k,1,N)REP(i,1,N){int dist=abs(X[k]-i)+abs(Y[k]-i);w[k][i]=-dist;}KM();int minn=0;REP(i,1,N)minn+=-w[Left[i]][i];ans=min(ans,minn);REP(k,1,N)REP(i,1,N){int dist=abs(X[k]-i)+abs(Y[k]-N+i-1);w[k][i]=-dist;}KM();minn=0;REP(i,1,N)minn+=-w[Left[i]][i];ans=min(ans,minn);printf("Board %d: %d moves required.\n\n",++cas,ans);}return 0;
}