Building roads

Building roads

Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 34 Accepted Submission(s): 13

Problem Description

Farmer John's farm has N barns, and there are some cows that live in each barn. The cows like to drop around, so John wants to build some roads to connect these barns. If he builds roads for every pair of different barns, then he must build N * (N - 1) / 2 roads, which is so costly that cheapskate John will never do that, though that's the best choice for the cows.

Clever John just had another good idea. He first builds two transferring point S1 and S2, and then builds a road connecting S1 and S2 and N roads connecting each barn with S1 or S2, namely every barn will connect with S1 or S2, but not both. So that every pair of barns will be connected by the roads. To make the cows don't spend too much time while dropping around, John wants to minimize the maximum of distances between every pair of barns.

That's not the whole story because there is another troublesome problem. The cows of some barns hate each other, and John can't connect their barns to the same transferring point. The cows of some barns are friends with each other, and John must connect their barns to the same transferring point. What a headache! Now John turns to you for help. Your task is to find a feasible optimal road-building scheme to make the maximum of distances between every pair of barns as short as possible, which means that you must decide which transferring point each barn should connect to.

We have known the coordinates of S1, S2 and the N barns, the pairs of barns in which the cows hate each other, and the pairs of barns in which the cows are friends with each other.

Note that John always builds roads vertically and horizontally, so the length of road between two places is their Manhattan distance. For example, saying two points with coordinates (x1, y1) and (x2, y2), the Manhattan distance between them is |x1 - x2| + |y1 - y2|.

 

Input

The first line of input consists of 3 integers N, A and B (2 <= N <= 500, 0 <= A <= 1000, 0 <= B <= 1000), which are the number of barns, the number of pairs of barns in which the cows hate each other and the number of pairs of barns in which the cows are friends with each other.

Next line contains 4 integer sx1, sy1, sx2, sy2, which are the coordinates of two different transferring point S1 and S2 respectively.

Each of the following N line contains two integer x and y. They are coordinates of the barns from the first barn to the last one.

Each of the following A lines contains two different integers i and j(1 <= i < j <= N), which represent the i-th and j-th barns in which the cows hate each other.

The same pair of barns never appears more than once.

Each of the following B lines contains two different integers i and j(1 <= i < j <= N), which represent the i-th and j-th barns in which the cows are friends with each other. The same pair of barns never appears more than once.

You should note that all the coordinates are in the range [-1000000, 1000000].

 

Output

You just need output a line containing a single integer, which represents the maximum of the distances between every pair of barns, if John selects the optimal road-building scheme. Note if there is no feasible solution, just output -1.

 

Sample Input

4 1 1
12750 28546 15361 32055
6706 3887
10754 8166
12668 19380
15788 16059
3 4
2 3

 

Sample Output

53246

 

Source

POJ Monthly - 2006.01.22 - zhucheng

 

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/*
题意：有n个牛棚，给出坐标，农夫想先建两个中转站s1,s2，然后每个牛棚通过中转站进行相互联通，但是给出a,b牛棚的牛相互厌恶
    不能通过同一个中转站，c,d两个牛棚的牛相互喜欢，必须通过同一个中转站。s1 ,s2是连通的中间有一条路，现在让你求怎么样
    建边，才能使这些牛棚距离最大的两个牛棚的距离最小。
 
题意：和maximum shortest distance（最大团）相似，就是二分距离建边，然后判断的时候只需要用2-SAT跑一下看是否可以解决就可
    以,问题的关键就在于如何建边，每次二分判断的时候，先按照A B的要求进行建边，然后按照如果距离大于mid的建边，然后再判
    段是不是可以解决2-SAT问题。
 
#错误：RE正在debug
       build 写的不是很好
*/
#include<bits/stdc++.h>
using namespace std;
/*********************************************2-SAT模板*********************************************/
const int maxn=+;
struct TwoSAT
{
    int n;//原始图的节点数(未翻倍)
    vector<int> G[maxn*];//G[i].j表示如果mark[i]=true,那么mark[j]也要=true
    bool mark[maxn*];//标记
    int S[maxn*],c;//S和c用来记录一次dfs遍历的所有节点编号
 
    //从x执行dfs遍历,途径的所有点都标记
    //如果不能标记,那么返回false
    bool dfs(int x)
    {
        if(mark[x^]) return false;//这两句的位置不能调换
        if(mark[x]) return true;
        mark[x]=true;
        S[c++]=x;
        for(int i=;i<G[x].size();i++)
            if(!dfs(G[x][i])) return false;
        return true;
    }
 
    void init(int tol)
    {
        n=tol;
        for(int i=;i<*tol;i++)
            G[i].clear();
        memset(mark,,sizeof(mark));
    }
 
    //加入(x,xval)或(y,yval)条件
    //xval=0表示假，yval=1表示真
    void add_clause(int x,int xval,int y,int yval)//这个地方不是一尘不变的，而是参照问题的约束条件进行加边
    {
        x=x*+xval;
        y=y*+yval;
        G[x^].push_back(y);//这是建双向边
        G[y^].push_back(x);
    }
 
    //判断当前2-SAT问题是否有解
    bool solve()
    {
        for(int i=;i<*n;i+=)
        if(!mark[i] && !mark[i+])
        {
            c=;
            if(!dfs(i))
            {
                while(c>) mark[S[--c]]=false;
                if(!dfs(i+)) return false;
            }
        }
        return true;
    }
}TS;
/*********************************************2-SAT模板*********************************************/
struct Point{
    int x,y;
    Point(){}
    Point(int a,int b){
        x=a;
        y=b;
    }
    void input(){
        scanf("%d%d",&x,&y);
    }
};
int dis(Point a,Point b){//曼哈顿距离
    int dx=a.x-b.x;
    int dy=a.y-b.y;
    return abs(dx)+abs(dy);
}
int n,A,B;
Point s1,s2;//中转站
Point p[maxn];//牛棚的坐标
Point a[maxn*],b[maxn*];//用来标记A B给出的关系
int g[maxn][maxn][];//离散化两点间的距离，两点的距离总共有四种状态，都在s1,都在s2，交叉的两种
int sTos=;//s1和s2间的距离
 
void init(){//初始化出所有的两点间的距离
    for(int i=;i<n;i++){
        for(int j=;j<i;j++){
            g[i][j][]=dis(p[i],s1)+dis(p[j],s1);//都在s1
            g[i][j][]=dis(p[i],s1)+dis(p[j],s2)+sTos;//i在s1 j在s2
            g[i][j][]=dis(p[i],s2)+dis(p[j],s1)+sTos;//i在s2 j在s1
            g[i][j][]=dis(p[i],s2)+dis(p[j],s2);//都在s2
        }
    }
}
 
bool judge(int mid){//按照要求将所有的边建好
    TS.init(n);                                                      /* × */
    for(int i=;i<A;i++){//相互喜欢的，都在s1或者s2
        TS.add_clause(a[i].x-,,a[i].y-,);
        TS.add_clause(a[i].y-,,a[i].x-,);
    }
    for(int i=;i<B;i++){//相互讨厌的，只要不在一块就行
        TS.add_clause(b[i].x-,,b[i].x-,);
        TS.add_clause(b[i].y-,,b[i].y-,);
        TS.add_clause(b[i].x-,,b[i].y-,);
        TS.add_clause(b[i].y-,,b[i].x-,);
    }
 
    for(int i=;i<n;i++){                                              /* √ */
        for(int j=;j<i;j++){
            if(g[i][j][]>mid){
                TS.add_clause(i,,j,);//都在s1
            }
            if(g[i][j][]>mid){
                TS.add_clause(i,,j,);//i在s1 j在s2
            }
            if(g[i][j][]>mid){
                TS.add_clause(i,,j,);//i在s2 j在s1
            }
            if(g[i][j][]>mid){
                TS.add_clause(i,,j,);//都在s2
            }
        }
    }
    return TS.solve();
}
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d%d%d",&n,&A,&B)!=EOF){
        s1.input();s2.input();
        // cout<<s1.x<<" "<<s1.y<<" "<<s2.x<<" "<<s2.y<<endl;
        sTos=dis(s1,s2);
        // cout<<sTos<<endl;
        for(int i=;i<n;i++){
            p[i].input();
            // cout<<p[i].x<<" "<<p[i].y<<endl;
        }//处理点的输入
        init();
        // for(int i=0;i<n;i++){
            // for(int j=0;j<n;j++){
                // cout<<g[i][j][0]<<" ";
            // }cout<<endl;
        // }
        for(int i=;i<A;i++){
            a[i].input();
            // cout<<a[i].x<<" "<<a[i].y<<endl;
        }
        for(int i=;i<B;i++){
            b[i].input();
            // cout<<b[i].x<<" "<<b[i].y<<endl;
        }
        if(judge()==false){//如果这种状态不可能的就直接输出就行了
            puts("-1");
            continue;
        }
        int l=,r=;
        while(l<r){
            // cout<<l<<" "<<r<<endl;
            int mid=(l+r)/;
            if(judge(mid)==false)
                l=mid+;
            else r=mid;
        }
        printf("%d\n",l);
    }
    return ;
}