POJ 1723 SOLDIERS (水)

Description

给出士兵的位置（x, y）让士兵排成一排，计算士兵移动的最小的距离

Sample Input

5
1 2
2 2
1 3
3 -2
3 3

Sample Output

8

贪心
先将他们移动到同一横排，这横排应该是他们纵坐标的中位数，才能使得此过程总步数最少。然后要把他们紧凑起来。我们先把横坐标排序，并假设起点是a，那么我们就是要求abs(a+i - xi),i=1~n的加和。即abs(a-(xi-i)),i=1~n的加和。我们构建一个新数列zi = xi - i，当a等于z的中位数时原式的值最小

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;const int MAX = 10010;int x[MAX];int y[MAX];int z[MAX];int main(){    int N;    while (~scanf("%d", &N))    {        memset(x, 0, sizeof(x));        memset(y, 0, sizeof(x));        memset(z, 0, sizeof(x));        for (int i = 0; i < N; i++)        {            scanf("%d%d", &x[i], &y[i]);        }        sort(x, x + N);        sort(y, y + N);        for (int i = 0; i < N; i++)        {            z[i] = x[i] - i;        }        sort(z, z + N);        int ans_x = z[N / 2], ans_y = y[N / 2], res = 0;        for (int i = 0; i < N; i++)        {            res += abs(z[i] - ans_x);            res += abs(y[i] - ans_y);        }        printf("%d", res);    }    return 0;}