# BZOJ3170: [Tjoi2013] squirrel Party (Chebyshev distance to Manhattan)

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 1524  Solved: 803
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## Description

There are N little squirrels. Their household uses a point x,y to express that the distance between the two points is defined as: the point (x,y) and the eight points around it, that is, four points up, down, left, right, and four points diagonally. The distance is 1. Now N squirrels have to go to a squirrels' house to ask for the shortest distance.

## Input

The first line gives the number n, indicating how many squirrels there are. 0<=N<=10^5
Next N lines, each line gives x,y the coordinates of its home.
-10^9<=x,y<=10^9

## Output

Indicates the distance and minimum for the party.

6
-4 -1
-1 -2
2 -4
0 2
0 3
5 -2

20

## Source

emmm, the topic gives the Chebyshev distance. We need to convert it into Manhattan distance

For each point $(x, y)$in the coordinates, it is converted to Manhattan distance followed by $(\ frac{x + y}{2}, \frac{x - y}{2})$

Then enumerate a point of $k$. What we need to calculate is $\ sum {I = 1} ^ n | x | K - x | + | y | K - y ||$

After brute force splitting, we found that prefixes and optimizations can be used

The code is very beautiful.

#include<cstdio>
#include<algorithm>
#define LL long long
using namespace std;
const int MAXN = 1e5 + 10;
const LL INF = 1e18 + 10;
char c = getchar();int x = 0,f = 1;
while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = x * 10 + c - '0',c = getchar();}
return x * f;
}
LL N, x[MAXN], y[MAXN], sortx[MAXN], sorty[MAXN], sumx[MAXN], sumy[MAXN];
LL QueryX(int l, int r) {
return sumx[r] - sumx[l - 1];
}
LL QueryY(int l, int r) {
return sumy[r] - sumy[l - 1];
}
LL calc(LL k) {
LL posx = lower_bound(sortx + 1, sortx + N + 1, x[k]) - sortx,
posy = lower_bound(sorty + 1, sorty + N + 1, y[k]) - sorty;
return posx * sortx[posx] - QueryX(1, posx) - (N - posx) * sortx[posx] + QueryX(posx + 1, N) +
posy * sorty[posy] - QueryY(1, posy) - (N - posy) * sorty[posy] + QueryY(posy + 1, N);
}
int main() {
#ifdef WIN32
freopen("a.in", "r", stdin);
#endif
for(int i = 1; i <= N; i++) {
x[i] = sortx[i] = a + b,
y[i] = sorty[i] = a - b;
}
sort(sortx + 1, sortx + N + 1);
sort(sorty + 1, sorty + N + 1);
for(int i = 1; i <= N; i++)
sumx[i] = sumx[i - 1] + sortx[i],
sumy[i] = sumy[i - 1] + sorty[i];
LL ans = INF;
for(int i = 1; i <= N; i++)
ans = min(ans, calc(i));
printf("%lld", ans >> 1);
return 0;
}

Tags: C++

Posted on Thu, 13 Feb 2020 16:25:24 -0500 by lucifersolutions