BZOJ3170: [Tjoi2013]松鼠聚会

描述

有N个小松鼠，它们的家用一个点x,y表示，两个点的距离定义为：点(x,y)和它周围的8个点即上下左右四个点和对角的四个点，距离为1。现在N个松鼠要走到一个松鼠家去，求走过的最短距离。

题解

简直就是个高中数学题啊。。。蒟蒻数学不好，学不来呜

松鼠$i $ 和 $j$ 的距离为$ max{leftvert X_i - X_j ightvert, leftvert Y_i - Y_j ightvert } $

但是我们不能在很短的时间内求出对于所有$i$和$j$ 的$max$， 所以只能通过加绝对值让$max$去掉。

首先1.  $max{a, b} = (leftvert a + b ightvert + leftvert a - b ightvert) div 2$

并且 2.$ leftvert leftvert a ightvert - leftvert b ightvert ightvert + leftvert leftvert a ightvert  + leftvert b ightvert ightvert  = leftvert a - b ightvert + leftvert a + b ightvert$

我们通过式子1算出$ max{leftvert X_i - X_j ightvert, leftvert Y_i - Y_j ightvert }  = ( leftvert leftvert  X_i - X_j ightvert - leftvert Y_i - Y_j ightvert ightvert + leftvert leftvert X_i - X_j ightvert  + leftvert Y_i - Y_j ightvert ightvert ) div 2$

然后再通过2式就变成了 $( leftvert ( X_i-Y_i ) - (X_j - Y_j) ightvert + leftvert (X_i - Y_i)  + (X_j - Y_j) ightvert) div 2$

令$a_i = X_i - Y_i $ $b_i = X_i + Y_i$

算出所有的$leftvert a_i  - a_j ightvert $ 和$leftvert b_i - b_j ightvert$ 再除 2， 取答案最小的$i$

算$leftvert a_i  - a_j ightvert $ 时需要排序求前缀和计算。

打绝对值累死我了。。、

代码

#include<cstring>
#include<cstdio>
#include<algorithm>
#define rd read()
#define ll long long
using namespace std;

const int N = 2e5;

ll sumx[N], sumy[N], n, ans = 1e18;

struct node {
    ll x, y;
    ll xx, yy;
    ll disx, disy;
}a[N];

ll read() {
    ll X = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}

int cmpx(const node &A, const node &B ) {
    return A.xx < B.xx;
}

int cmpy(const node &A, const node &B) {
    return A.yy < B.yy;
}

int main()
{
    n = rd;
    for(int i = 1; i <= n; ++i) {
        a[i].x = rd * 2;
        a[i].y = rd * 2;
        a[i].xx = (a[i].x + a[i].y) >> 1;
        a[i].yy = (a[i].x - a[i].y) >> 1;
    }
    sort(a+1, a+1+n, cmpx);
    for(int i = 1; i <= n; ++i) sumx[i] = sumx[i - 1] + a[i].xx;
    for(int i = 1; i <= n; ++i) {
        a[i].disx = a[i].xx * i - sumx[i];
        a[i].disx += sumx[n] - sumx[i] - (n - i) * a[i].xx;
    }
    sort(a+1, a+1+n, cmpy);
    for(int i = 1; i <= n; ++i) sumy[i] = sumy[i - 1] + a[i].yy;
    for(int i = 1; i <= n; ++i) {
        a[i].disy = a[i].yy * i - sumy[i];
        a[i].disy += sumy[n] - sumy[i] - (n - i) * a[i].yy;
        if(a[i].disy + a[i].disx < ans) ans = a[i].disx + a[i].disy;
    }
    printf("%lld
", ans >> 1);
}