BZOJ 2241 打地鼠(特技暴力)

果然暴力出奇迹。。 O(n^2m^2)=1e8 536ms能过。

枚举锤子的长和宽，再验证是否可以满足条件并更新答案。

我们先从左上角为(1,1)的先锤，显然锤的次数是a[1][1]. 锤(i,j)的时候呢，算一下右下角为(i,j)的锤数组的矩形面积，然后更新(i,j)的值。

用二维前缀和可以做到O(1).

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

int a[N][N], sum[N][N];

int get_sum(int x, int y, int n, int m){
    return sum[x][y]-(x>=n?sum[x-n][y]:)-(y>=m?sum[x][y-m]:)+(x>=n&&y>=m?sum[x-n][y-m]:);
}
int main ()
{
    int n, m, ans=INF, flag;
    scanf("%d%d",&n,&m);
    FOR(i,,n) FOR(j,,m) scanf("%d",&a[i][j]);
    FOR(l,,n) FOR(r,,m) {
        flag=;
        FOR(i,,n) FOR(j,,m) {
            int d=get_sum(i,j-,l,r-)+get_sum(i-,j,l-,r)-get_sum(i-,j-,l-,r-);
            if (d>a[i][j]) {flag=-; break;}
            if ((i>n-l+||j>m-r+)&&d!=a[i][j]) {flag=-; break;}
            flag+=(a[i][j]-d);
            sum[i][j]=sum[i-][j]+sum[i][j-]-sum[i-][j-]+(a[i][j]-d);
        }
        if (flag!=-) ans=min(ans,flag);
    }
    printf("%d\n",ans);
    return ;
}