题面
Sol
把每个节点都加上(2^n-1)
那么非叶节点的编号就是(1)到(2^n-1)
就可以把它当成是一棵线段树了
然后看收费的方法
意思就是(A)多就收(A)的
否则收(B)的费用
可以处理出每个叶节点经过某个(LCA)的费用和
那么枚举这条链上的非叶节点的(A)是否大于(B)就可以算出要收的费用了
然后设(f[i][j])表示(i)号点，观察到的叶节点选了(j)个(A)的最小代价
假如枚举的是(A>B)那么背包转移时，要保证(j)大于区间长度的一半
否则
要保证小于一半
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
const int _(2050);
const int INF(1e9);
typedef long long ll;
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, pw[11], a[_], c[_], cost[_][_], f[_][_], ans = 2e9, cp[_];
IL int LCA(RG int x, RG int y){
while(x != y) x = x >> 1, y = y >> 1;
return x;
}
IL void Solve(RG int p, RG int d){
if(!d){
f[p][a[p]] = 0, f[p][a[p] ^ 1] = c[p];
for(RG int i = p >> 1; i; i >>= 1) f[p][cp[i]] += cost[i][p];
return;
}
for(RG int i = 0; i <= pw[n]; ++i) f[p][i] = INF;
cp[p] = 0, Solve(p << 1, d - 1), Solve(p << 1 | 1, d - 1);
for(RG int i = 1; i <= pw[d - 1]; ++i)
for(RG int j = pw[d - 1] - i + 1; i + j <= pw[d]; ++j)
f[p][i + j] = min(f[p][i + j], f[p << 1][i] + f[p << 1 | 1][j]);
cp[p] = 1, Solve(p << 1, d - 1), Solve(p << 1 | 1, d - 1);
for(RG int i = 0; i <= pw[d - 1]; ++i)
for(RG int j = 0; i + j <= pw[d - 1]; ++j)
f[p][i + j] = min(f[p][i + j], f[p << 1][i] + f[p << 1 | 1][j]);
}
int main(RG int argc, RG char* argv[]){
n = Input(), pw[0] = 1;
for(RG int i = 1; i <= 10; ++i) pw[i] = pw[i - 1] << 1;
for(RG int i = 1; i <= pw[n]; ++i) a[pw[n] + i - 1] = Input();
for(RG int i = 1; i <= pw[n]; ++i) c[pw[n] + i - 1] = Input();
for(RG int i = 1; i < pw[n]; ++i)
for(RG int j = i + 1; j <= pw[n]; ++j){
RG int x = i + pw[n] - 1, y = j + pw[n] - 1;
RG int v = Input(), d = LCA(x, y);
cost[d][x] += v, cost[d][y] += v;
}
Solve(1, n);
for(RG int i = 0; i <= pw[n]; ++i) ans = min(ans, f[1][i]);
printf("%d
", ans);
return 0;
}