UVA1045 The Great Wall Game

Link Title: UVA1045 of The Great Wall Game
Title effect: there are n pieces on a chessboard n * n, requires that the pieces piece by moving the arrangement of one of the following conditions: the form transverse ranks; form longitudinal rows; form of moldings arrangement (there are two).
Solution: questions of final status this small, so we can enumerate the ultimate end state, then calculate, as to how to calculate, with a maximum flow can also be a perfect match with the bipartite graph, I use the latter .
Code:

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int MAXNODE = 105;
typedef int Type;
const Type INF=0x3f3f3f3f;
struct KM{
    int n;
    Type g[MAXNODE][MAXNODE];
    Type Lx[MAXNODE],Ly[MAXNODE],slack[MAXNODE];
    int left[MAXNODE];
    bool S[MAXNODE],T[MAXNODE];
    void init(int n){
        this->n=n;
    }
    void add_Edge(int u,int v,Type val){
        g[u][v]=val;
    }
    bool dfs(int i){
        S[i]=1;
        for(int j=0;j<n;j++){
            if(T[j]){
                continue;
            }
            Type tmp=Lx[i]+Ly[j]-g[i][j];
            if(!tmp){
                T[j]=1;
                if(left[j]==-1||dfs(left[j])){
                    left[j]=i;
                    return 1;
                }
            }
            else{
                slack[j]=min(slack[j],tmp);
            }
        }
        return 0;
    }
    void update(){
        Type a=INF;
        for(int i=0;i<n;i++){
            if(!T[i]){
                a=min(a,slack[i]);
            }
        }
        for(int i=0;i<n;i++){
            if(S[i]){
                Lx[i]-=a;
            }
            if(T[i]){
                Ly[i]+=a;
            }
        }
    }
    Type km(){
        for(int i=0;i<n;i++){
            left[i]=-1;
            Lx[i]=-INF;
            Ly[i]=0;
            for(int j=0;j<n;j++){
                Lx[i]=max(Lx[i],g[i][j]);
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                slack[j]=INF;
            }
            while(1){
                for(int j=0;j<n;j++){
                    S[j]=T[j]=0;
                }
                if(dfs(i)){
                    break;
                }
                else{
                    update();
                }
            }
        }
        Type ans=0;
        for (int i=0;i<n;i++){
            ans+=g[left[i]][i];
        }
        return ans;
    }
}gao;
const int N=20;
int n,x[N],y[N];
int dis(int x1,int y1,int x2,int y2){
    return abs(x1-x2)+abs(y1-y2);
}
int main(){
    int cas=0;
    while(~scanf("%d",&n)&&n){
        gao.init(n);
        for(int i=0;i<n;i++){
            scanf("%d%d",&x[i],&y[i]);
            x[i]--;
            y[i]--;
        }
        int ans=-1000;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                for(int k=0;k<n;k++){
                    gao.add_Edge(j,k,-dis(x[j],y[j],i,k));
                }
            }
            ans=max(ans,gao.km());
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                for(int k=0;k<n;k++){
                    gao.add_Edge(j,k,-dis(x[j],y[j],k,i));
                }
            }
            ans=max(ans,gao.km());
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                gao.add_Edge(i,j,-dis(x[i],y[i],j,j));
            }
        }
        ans=max(ans,gao.km());
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                gao.add_Edge(i,j,-dis(x[i],y[i],n-j-1,j));
            }
        }
        ans=max(ans,gao.km());
        printf("Board %d: %d moves required.\n\n",++cas,-ans);
    }
    return 0;
}