[CF85E] Guard Towers

题目描述

In a far away kingdom lives a very greedy king. To defend his land, he built

During defence operations to manage a guard tower a general has to send part of his army to that tower. Each general asks some fee from the king for managing towers. As they live in a really far away kingdom, each general evaluates his fee in the following weird manner: he finds two remotest (the most distant) towers, where the soldiers of his army are situated and asks for the fee equal to the distance. Each tower is represented by a point on the plane with coordinates $(x_{1},y_{1}) (x1,y1) and (x2,y2) (x_{2},y_{2}) (x2,y2) is determined in this kingdom as ∣x1−x2∣+∣y1−y2∣ |x_{1}-x_{2}|+|y_{1}-y_{2}| ∣x1−x2∣+∣y1−y2∣ .$

The greedy king was not exactly satisfied with such a requirement from the generals, that's why he only agreed to pay one fee for two generals, equal to the maximum of two demanded fees. However, the king is still green with greed, and among all the ways to arrange towers between armies, he wants to find the cheapest one. Each tower should be occupied by soldiers of exactly one army.

He hired you for that. You should find the minimum amount of money that will be enough to pay the fees. And as the king is also very scrupulous, you should also count the number of arrangements that will cost the same amount of money. As their number can be quite large, it is enough for the king to know it as a remainder from dividing by $10^{9}+7 109+7 .$

Two arrangements are distinct if the sets of towers occupied by soldiers of the first general are distinct.

输入输出格式

输入格式：

The first line contains an integer

Pretest 6 is one of the maximal tests for this problem.

输出格式：

Print on the first line the smallest possible amount of money that will be enough to pay fees to the generals.

Print on the second line the number of arrangements that can be carried out using the smallest possible fee. This number should be calculated modulo $10^{9}+7 109+7 ).$

输入输出样例

输入样例#1：

2
0 0
1 1

输出样例#1：

0
2

输入样例#2：

输出样例#2：

1
4

输入样例#3：

输出样例#3：

2000
2

说明

In the first example there are only two towers, the distance between which is equal to 2. If we give both towers to one general, then we well have to pay 2 units of money. If each general receives a tower to manage, to fee will be equal to 0. That is the smallest possible fee. As you can easily see, we can obtain it in two ways.

二分判断是否是二分图；

二分集合里的最小值，把大于这个值的连上一条边，然后判断是否是二分图；

我因为没开long long调了一上午，生无可恋。

Code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
const int mod = 1000000007;

int n, X[5005], Y[5005];
int dis[5005][5005];
int color[5005];
int l = 0 , r=1e6;
int bel[5005];

inline long long ksm(long long a, long long y)
{
    int res = 1;
    while (y)
    {
        if (y & 1) res = (res % mod * a % mod) % mod;
        a = (a % mod * a % mod) % mod;
        y >>= 1;
    }
    return res % mod;
}

bool dfs(int x, int col, int minn)
{
    color[x] = col;
    for (register int i = 1 ; i <= n ; i ++)
    {
        if (x == i) continue;
        if (dis[x][i] > minn)
        {
            if (color[i] == -1)
            {
                if (!dfs(i, 3-col, minn)) return 0;
            }
            else if (color[i] != 3 - col) return 0;
        }
    }
    return 1;
}

inline bool check(int x)
{
    memset(color, -1, sizeof color);
    for (register int i = 1 ; i <= n ; i ++)
        if (color[i] == -1)
            if (!dfs(i, 1, x)) return 0;
    return 1;
}

void dfs2(int x, int minn, int nu)
{
    bel[x] = nu;
    for (register int i = 1 ; i <= n ; i ++)
    {
        if (i == x) continue;
        if (bel[i]) continue;
        if (dis[x][i] <= minn) continue;
        dfs2(i, minn, nu);
    }
}

int main()
{
    scanf("%d", &n);
    for (register int i = 1 ; i <= n ; i ++) scanf("%d%d",&X[i],&Y[i]);
    for (register int i = 1 ; i <= n ; i ++)
        for (register int j = 1 ; j <= n ; j ++)
            dis[i][j] = abs(X[i]-X[j]) + abs(Y[i]-Y[j]);
    int ans;
    while (l < r)
    {
        int mid = l + r >> 1;
        if (check(mid)) r = mid;
        else l = mid + 1;
    }
    ans = l;
    cout << ans << endl;
    int tot = 0;
    for (register int i = 1 ; i <= n ; i ++)
    {
        if (!bel[i]) tot++, dfs2(i, ans, tot);
    }
    cout << (ksm(2, tot)) % mod;
    return 0;
}