HDU 3687 Brute Force

In the N * M matrix, N * n people are distributed, N rows are arranged, and only left and right movements are allowed. The minimum cost required to merge all people into N * n squares is calculated.

Because each person can only move in the row, every merge method of each row is pre-processed, and then the column's

#include "stdio.h"#include "string.h"#include "iostream"#include "algorithm"using namespace std;int inf=0x3f3f3f3f;int Fabs(int a){    if (a<0) return -a; else return a;}int Min(int a,int b){    if (a<b) return a;    else return b;}struct node{    int num;    int x[210];}mark[210];int main(){    int n,m,i,j,k,sum,ans;    int dp[210][210];    while (scanf("%d%d",&n,&m)!=EOF)    {        if (n==0 && m==0) break;        memset(mark,0,sizeof(mark));        for (i=1;i<=n*n;i++)        {            scanf("%d%d",&j,&k);            mark[j].num++;            mark[j].x[mark[j].num]=k;        }        for (i=1;i<=n;i++)            sort(mark[i].x+1,mark[i].x+1+n);        memset(dp,0,sizeof(dp));        for (i=1;i<=n;i++)            for (j=1;j<=m-n+1;j++)                for (k=1;k<=n;k++)                    dp[i][j]+=Fabs(mark[i].x[k]-j-k+1);        ans=inf;        for (j=1;j<=m-n+1;j++)        {            sum=0;            for (i=1;i<=n;i++)                sum+=dp[i][j];            ans=Min(sum,ans);        }        printf("%d\n",ans);    }    return 0;}

HDU 3687 Brute Force