bzoj4776[Usaco2017 Open]Modern Art

题目意思就不说了，挺明显的。

想法也挺明显的，就是把每个数字最大出现矩形都框起来，然后看那些格子的出现次数>1，这个格子里面的数字肯定不是第一个，然后去掉就好了。
框格子用二维前缀和就好了
1A.

#include<cstdio>#include<algorithm>#include<iostream>#include<queue>#include<cstring>#define fo(i,a,b) for(int i=a;i<=b;i++)#define fd(i,a,b) for(int i=a;i>=b;i--)#define inf 0x3f3f3f3fusing namespace std;const int N=1e3+5;const int M=N*N;int n,cnt,ans;int a[N][N],tim[N][N];bool vis[M];int up[M],down[M],right1[M],left1[M];int main(){    scanf("%d",&n);    memset(up,0x3f,sizeof(up));    memset(left1,0x3f,sizeof(left1));    fo(i,1,n)    fo(j,1,n)    {        scanf("%d",&a[i][j]);        if (a[i][j])        {            if (up[a[i][j]]==inf)++cnt;            up[a[i][j]]=min(up[a[i][j]],i);            down[a[i][j]]=max(down[a[i][j]],i);            left1[a[i][j]]=min(left1[a[i][j]],j);            right1[a[i][j]]=max(right1[a[i][j]],j);        }    }    fo(i,1,n*n)    if (up[i]!=inf)    {        tim[up[i]][left1[i]]++;        tim[down[i]+1][left1[i]]--;        tim[up[i]][right1[i]+1]--;        tim[down[i]+1][right1[i]+1]++;    }    /*fo(i,1,n)    {        fo(j,1,n)        printf("%d ",tim[i][j]);        printf("\n");    }*/    fo(i,1,n)    fo(j,1,n)    tim[i][j]+=tim[i-1][j]+tim[i][j-1]-tim[i-1][j-1];    fo(i,1,n)    fo(j,1,n)    if (a[i][j]&&tim[i][j]>1)vis[a[i][j]]=1;    fo(i,1,n*n)if (!vis[i])ans++;    if (cnt==1&&n!=1)--ans;    printf("%d\n",ans);    return 0;}