# [NOI2006] network charge

An interesting question for thieves. Think about converting the cost into something. Watch If we define the state of leaf node {A,0},{B,1}}, the state of non leaf node {Na > = Nb, 0}, {Na < Nb, 1}}, we can get the conclusion by combining the above figure

• The states of leaf nodes x, y (x < y) are the same
• The state of leaf node is the same as lca(x,y), and the cost is 0
• The state of leaf node is different from that of lca(x,y), and the cost is 2f[x,y]
• The states of leaf nodes x, y (x < y) are different, the cost f[x,y]

Technically, the weight of leaf node x about ancestor D is directly defined as H (x, d) = [state of X ≠ state of D] ∑ f[x,y], where y is the leaf node not equal to x governed by D subtree.

In this way, if the state of ancestors is clear, the contribution of X is equal to Σ h(x,d).

Then we consider enumerating the states of ancestors and dp, and set f[x,i] to represent the minimum cost of i 0 state leaf nodes in the X subtree. The transition part is relatively simple, and the reference code is OK.

I didn't make it clear that I will make up tomorrow. Remember to initialize dp array!!!

```#include <bits/stdc++.h>
#define ls (x<<1)
#define rs (x<<1|1)
using namespace std;
const int inf=0x3f3f3f3f;

int n;
int cv,ori,tmp;
int dp,v;
int lq,rq;

void dfs(int x,int l,int r,int set,int dep) {
memset(dp[x],inf,sizeof dp[x]);
if(l==r) {
dep--;
dp[x]=ori[l]?0:cv[l];
dp[x]=ori[l]?cv[l]:0;
for(int i=1; i<=dep; ++i) {
int mid=(lq[i]+rq[i])>>1;
int key=!(1&(set>>(dep-i))); //Different contributions
if(l<=mid) dp[x][key]+=v[l][rq[i]]-v[l][mid];
else dp[x][key]+=v[l][mid]-v[l][lq[i]-1];
}
return;
}
int mid=(l+r)>>1,len=r-l+1;
lq[dep]=l,rq[dep]=r;

dfs(ls,l,mid,set<<1,dep+1);
dfs(rs,mid+1,r,set<<1,dep+1);

for(int i=0; i<len/2; ++i)
for(int j=0; j<=i; ++j)
dp[x][i]=min(dp[x][i],dp[ls][j]+dp[rs][i-j]);

dfs(ls,l,mid,set<<1|1,dep+1);
dfs(rs,mid+1,r,set<<1|1,dep+1);
for(int i=len/2; i<=len; ++i)
for(int j=0; j<=i; ++j)
dp[x][i]=min(dp[x][i],dp[ls][j]+dp[rs][i-j]);
}

int main() {
scanf("%d",&n); n=1<<n;
for(int i=1; i<=n; ++i) scanf("%d",ori+i);
for(int i=1; i<=n; ++i) scanf("%d",cv+i);
for(int i=1; i<=n; ++i)
for(int j=i+1; j<=n; ++j) {
scanf("%d",&v[i][j]);
v[j][i]=v[i][j];
}
for(int i=1; i<=n; ++i)
for(int j=1; j<=n; ++j)
v[i][j]+=v[i][j-1];
dfs(1,1,n,0,1);
int ans=inf;
for(int i=0; i<=n; ++i)
ans=min(ans,dp[i]);
printf("%d\n",ans);
return 0;
}```

Tags: C++

Posted on Tue, 05 Nov 2019 14:33:47 -0500 by meandrew