跳跃式LIS(nlogn),在普通的转移基础上增加一种可以跨越一段距离的转移,用一颗新的树状数组维护,同时,我们还要维护跨越完一次后面的转移,所以我用了3颗树状数组。。

比赛的时候一句话位置写错了,然后就。。。雪崩

呆马:

 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
int a[][];
int sum,ans,num;
int t,n,m,x,y;
int dp[][][];
int main()
{
scanf("%d",&t);
for (int k=;k<=t;k++)
{
memset(a,,sizeof(a));
memset(dp,,sizeof(dp)); scanf("%d%d",&n,&m);
for(int i=;i<=n;i++)
for (int j=;j<=m;j++)
scanf("%d",&a[i][j]); dp[][][a[][]]=a[][]*a[][]; for (int i=;i<=n;i++)
{
for (int j=;j<=m;j++)
{
x=i-; y=j-;
if (i== && j==) continue;
for (int k=;k<=;k++)
{
if (dp[x][j][k]==) continue;
sum=k+a[i][j];
num=dp[x][j][k]+a[i][j]*a[i][j];
if (dp[i][j][sum]==) dp[i][j][sum]=num;
dp[i][j][sum]=min(dp[i][j][sum],num);
}
for (int k=;k<=;k++)
{
if (dp[i][y][k]==) continue;
sum=k+a[i][j];
num=dp[i][y][k]+a[i][j]*a[i][j];
if (dp[i][j][sum]==) dp[i][j][sum]=num;
dp[i][j][sum]=min(dp[i][j][sum],num);
} }
} ans=;
for (int i=;i<=;i++)
{
if (dp[n][m][i]==) continue;
num=(n+m-)*dp[n][m][i]-i*i;
//cout<<i<<' '<<dp[n][m][i]<<' '<<num<<endl;
ans=min(ans,num);
}
printf("Case #%d: %d\n",k,ans);
}
}

Removed Interval

04-07 17:09