Building roads

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 822    Accepted Submission(s): 239

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

这条题卡了好久呀 -。-

卡完构图卡空间， 卡完空间卡二分范围 ， 右区间要去到 800W ， 500W返WA 。

卡完二分范围卡编译器问题 ，

不知道为啥  ， 把run() 部分写在 主函数交 C++会超时， 交G++就过。

条题其实很入门 ， 依然是 2分 + two-sat ....

处理距离牛与牛之间的距离那里也先大好表，避免超时吧~

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <map>

using namespace std;
typedef long long LL;

const int N = 1010;
const int M = 1000010;
const int MAX = 8000000;

int n , A , B  , st[N] , top , dis_s1_s2 ;
int eh[N] , et[M] , nxt[M] , tot ;

bool mark[N];

struct node
{
    int x , y ;
}e[N] , h[N<<1] , s[2] ;

int d[510][3] ;

void init()
{
    tot = 0;
    memset ( eh , -1 ,sizeof eh );
    memset ( mark , false , sizeof mark );
}

void addedge( int u , int v)
{
    et[tot] = v , nxt[tot] = eh[u] , eh[u] = tot ++ ;
    et[tot] = u , nxt[tot] = eh[v] , eh[v] = tot ++ ;
}
//--------------------

inline int dis( int x1 , int y1 , int x2 , int y2 ){ return abs( x1 - x2 ) + abs( y1 - y2 ); }

bool dfs( int u )
{
    if( mark[u] ) return true ;
    if( mark[u^1] ) return false ;
    mark[u] = true;
    st[top++]  = u ;
    for( int i = eh[u] ; ~i ; i = nxt[i] ){
        int v = et[i];
        if( !dfs( v^1 ) ) return false ;
    }
    return true ;

}

bool solve( )
{
    for( int i = 0 ; i < 2 * n ; i += 2 ){
        if( !mark[i] && !mark[i+1] ){
            top = 0 ;
            if( !dfs( i ) ){
                while( top > 0 ) mark[ st[--top] ] = false ;
                if( !dfs( i + 1 ) ) return false ;
            }
        }
    }
    return true;
}

bool test( double DIS )
{
    init();

    for( int i = 0 ; i < A + B ; ++i ){
        if( i < A ) addedge( 2*h[i].x , 2*h[i].y ) , addedge( 2*h[i].x^1 , 2*h[i].y^1 ) ;
        else addedge( 2*h[i].x^1 , 2*h[i].y ) , addedge( 2*h[i].x , 2*h[i].y^1 ) ;
    }

    for( int i = 0 ; i < n ; i ++ ){
        for( int j = i + 1 ; j < n ; j ++ ){
            if(  d[i][0] + d[j][0] > DIS  ) addedge( 2*i , 2*j );
            if(  d[i][1] + d[j][0] + dis_s1_s2 > DIS  ) addedge( 2*i^1 , 2*j );
            if(  d[i][0] + d[j][1] + dis_s1_s2 > DIS  ) addedge( 2*i , 2*j^1 );
            if(  d[i][1] + d[j][1] > DIS  ) addedge( 2*i^1 , 2*j^1 );
        }
    }
    return solve();
}

void run()
{
    int x , y ;

    for( int i = 0 ; i < 2 ; ++i )
        scanf("%d%d",&s[i].x,&s[i].y);

    dis_s1_s2 = dis( s[0].x,s[0].y,s[1].x,s[1].y );

    for( int i = 0 ; i < n ; ++i ){
        scanf("%d%d",&x,&y );
        d[i][0] = dis( x , y , s[0].x , s[0].y );
        d[i][1] = dis( x , y , s[1].x , s[1].y );
    }

    for( int i = 0 ; i < A + B ; ++i ){
        scanf("%d%d",&h[i].x,&h[i].y);
        h[i].x -- ;h[i].y -- ;
    }

    int ans = -1 , l = 0 , r = MAX ;

    while( l <= r )
    {
        int mid = ( l + r ) >> 1 ;
        if( test(mid) )
            ans = mid , r = mid - 1 ;
        else
            l = mid + 1 ;
    }
    printf("%d\n",ans);
}

int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif
    while( scanf("%d%d%d",&n,&A,&B) != EOF )run();
}