[HDU3687 National Day Parade]

[关键字]：枚举+贪心

[题目大意]：给出N*M的矩阵和N*N个点，所有点只能左右移动，问将所有点排列成矩形的最小步数。

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[分析]：枚举所构成的矩形的左边界，然后同一行上的每个点一定是按从y值从小到大往枚举的矩形里走，所以先将所有点按x为第一关键字y为第二关键字排序，然后按顺序移入枚举的矩形并计算出步数求最小。

[代码]：

View Code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN=11210;
const int INF=0x7fffffff;

struct node
{
    int x,y;
}a[MAXN];
int N,M,ans;

int Calc(int sy)
{
    int sum=0;
    for (int i=1;i<=N;i++)
        for (int j=0;j<N;j++)
            sum+=abs(a[(i-1)*N+j+1].y-(sy+j));
    return sum;
}

bool cmp(node a,node b)
{
    if ((a.x<b.x) || ((a.x==b.x) && (a.y<b.y))) return 1;
    else return 0;
}

int main()
{
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    while (scanf("%d%d",&N,&M))
    {
        if (!N && !M) break;
        for (int i=1;i<=N*N;i++) scanf("%d%d",&a[i].x,&a[i].y);
        sort(a+1,a+N*N+1,cmp);
        //for (int i=1;i<=N*N;i++) printf("%d %d\n",a[i].x,a[i].y);
        ans=INF;
        for (int i=1;i<=M-N+1;i++)
        {
            int temp=Calc(i);
            //printf("%d %d\n",i,temp);
            ans=min(ans,temp);
        }
        printf("%d\n",ans);
    }
    return 0;
}