# BZOJ1564 NOI2009二叉查找树（区间dp）

首先按数据值排序，那么连续一段区间的dfs序一定也是连续的。

将权值离散化，设f[i][j][k]为i到j区间内所有点的权值都>=k的最小代价，转移时枚举根考虑是否修改权值即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
#define N 75
int n,m,b[N],f[N][N][N],sum[N];
struct data
{
int x,y,z;
bool operator <(const data&a) const
{
return x<a.x;
}
}a[N];
int main()
{
#ifndef ONLINE_JUDGE
freopen("bzoj1564.in","r",stdin);
freopen("bzoj1564.out","w",stdout);
const char LL[]="%I64d\n";
#else
const char LL[]="%lld\n";
#endif
sort(a+1,a+n+1);sort(b+1,b+n+1);
for (int i=1;i<=n;i++) a[i].y=lower_bound(b+1,b+n+1,a[i].y)-b;
for (int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i].z;
memset(f,42,sizeof(f));
for (int i=1;i<=n;i++)
{
for (int k=1;k<=n;k++)
f[i][i-1][k]=0;
for (int k=1;k<=a[i].y;k++)
f[i][i][k]=a[i].z;
for (int k=a[i].y+1;k<=n;k++)
f[i][i][k]=a[i].z+m;
}
for (int k=1;k<=n;k++) f[n+1][n][k]=0;
for (int k=2;k<=n;k++)
for (int i=1;i<=n-k+1;i++)
{
int j=i+k-1;
for (int root=i;root<=j;root++)
{
f[i][j][a[root].y]=min(f[i][j][a[root].y],f[i][root-1][a[root].y]+f[root+1][j][a[root].y]+sum[j]-sum[i-1]);
for (int d=1;d<=n;d++)
f[i][j][d]=min(f[i][j][d],f[i][root-1][d]+f[root+1][j][d]+sum[j]-sum[i-1]+m);
}
for (int d=n;d>=1;d--) f[i][j][d]=min(f[i][j][d],f[i][j][d+1]);
}
cout<<f[1][n][1];
return 0;
}

posted @ 2018-10-11 02:20  Gloid  阅读(113)  评论(0编辑  收藏  举报