1. Title link:
Travel by car
2. The main idea of the topic:
The Chinese question is a bit long, so I won’t mislead everyone...
3. Analysis:
First preprocess the amount:
ga[i]: Starting from city i, the next city for small A.
gb[i]: Starting from city i, the next city for small B.
f [i] [j] [ k]: starting from the city j, k go ahead and take the
city after the number reaches the step.
da [i] [j] [ k]: starting from the city j, k go first, go
db [i] [j] [ k]: starting from the city j, k go first, go
After obtaining the above results, we can use the time complexity of the multiplication to find the distance traveled by small A and small B respectively.
The method of solving the above results is given below.
ga[i]: Starting from city i, the next city for small A.
gb[i]: Starting from city i, the next city for small B.
Enumerate city i in reverse order, dichotomy can find the first city whose height is greater than or equal to city i, just search left and right.
void init_g() { set < pair <ll, int> > st; st.emplace(make_pair(inf, 0)); st.emplace(make_pair(inf + 1, 0)); st.emplace(make_pair(-inf, 0)); st.emplace(make_pair(-inf - 1, 0)); for(int i = n; i >= 1; --i) { auto j = st.lower_bound(make_pair(h[i], i)); --j, --j; ll mn = inf, sm = inf; int mni, smi; for(int k = 0; k < 4; ++k) { auto p = *j; ll d = (ll)abs(p.first - h[i]); if(d < mn) { sm = mn, smi = mni; mn = d, mni = p.second; } else if(d < sm) { sm = d, smi = p.second; } ++j; } ga[i] = smi, gb[i] = mni; st.emplace(make_pair(h[i], i)); } }
f [i] [j] [ k]: starting from the city j, k go ahead and take the
i = 0 ： f [0] [j] [0] = ga [j], f [0] [j] [1] = gb [j];
i = 1：f[1][j][k] = f[0, f[0][j][0], 1 - k];
i > 1：f[i][j][k] = f[i - 1][f[i - 1][j][k]][k];
void init_f() { for(int j = 1; j <= n; ++j) { f[0][j][0] = ga[j]; f[0][j][1] = gb[j]; } for(int j = 1; j <= n; ++j) { for(int k = 0; k <= 1; ++k) { f[1][j][k] = f[0][f[0][j][k]][1 - k]; } } for(int i = 2; i < N; ++i) { for(int j = 1; j <= n; ++j) { for(int k = 0; k <= 1; ++k) { f[i][j][k] = f[i - 1][f[i - 1][j][k]][k]; } } } }
da [i] [j] [ k]: starting from the city j, k go first, go
db [i] [j] [ k]: starting from the city j, k go first, go
First define the function get_dist(i, j) = abs(h[i]-h[j]);
i = 0 ： da [0] [j] [0] = get_dist (j, ga [j]), da [0] [j] [1] = 0;
db[0][j][0] = 0, db[0][j][1] = get_dist(j, gb[j]);
i = 1 ： yes [1] [j] [k] = yes [0] [j] [k] + yes [0] [f [0] [j] [k]] [1 - k];
db[1][j][k] = db[0][j][k] + db[0][f[0][j][k]][1 - k];
i> 1 ： yes [i] [j] [k] = yes [i - 1] [j] [k] + yes [i - 1] [f [i - 1] [j] [k]] [k ];
db[i][j][k] = db[i - 1][j][k] + db[i - 1][f[i - 1][j][k]][k];
int get_dist(int i, int j) { return (int)abs(h[i] - h[j]); } void init_d() { for(int j = 1; j <= n; ++j) { da[0][j][0] = get_dist(j, ga[j]), da[0][j][1] = 0; db[0][j][1] = 0, db[0][j][1] = get_dist(j, gb[j]); } for(int j = 1; j <= n; ++j) { for(int k = 0; k <= 1; ++k) { da[1][j][k] = da[0][j][k] + da[0][f[0][j][k]][1 - k]; db[1][j][k] = db[0][j][k] + db[0][f[0][j][k]][1 - k]; } } for(int i = 2; i < N; ++i) { for(int j = 1; j <= n; ++j) { for(int k = 0; k <= 1; ++k) { da[i][j][k] = da[i - 1][j][k] + da[i - 1][f[i - 1][j][k]][k]; db[i][j][k] = db[i - 1][j][k] + db[i - 1][f[i - 1][j][k]][k]; } } } }
4. Code implementation:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int M = (int)1e5;
const int N = (int)17;
const ll inf = 0x3f3f3f3f3f3f3f3f;
int n;
int h[M + 5];
int ga[M + 5], gb[M + 5];
int f[N][M + 5][2];
ll da[N][M + 5][2], db[N][M + 5][2];
void read()
{
scanf("%d", &n);
for(int i = 1; i <= n; ++i)
scanf("%d", &h[i]);
}
void init_g()
{
set < pair <ll, int> > st;
st.emplace(make_pair(inf, 0));
st.emplace(make_pair(inf + 1, 0));
st.emplace(make_pair(-inf, 0));
st.emplace(make_pair(-inf - 1, 0));
for(int i = n; i >= 1; --i)
{
auto j = st.lower_bound(make_pair(h[i], i));
--j, --j;
ll mn = inf, sm = inf;
int mni, smi;
for(int k = 0; k < 4; ++k)
{
auto p = *j;
ll d = (ll)abs(p.first - h[i]);
if(d < mn)
{
sm = mn, smi = mni;
mn = d, mni = p.second;
}
else if(d < sm)
{
sm = d, smi = p.second;
}
++j;
}
ga[i] = smi, gb[i] = mni;
st.emplace(make_pair(h[i], i));
}
}
void init_f()
{
for(int j = 1; j <= n; ++j)
{
f[0][j][0] = ga[j];
f[0][j][1] = gb[j];
}
for(int j = 1; j <= n; ++j)
{
for(int k = 0; k <= 1; ++k)
{
f[1][j][k] = f[0][f[0][j][k]][1 - k];
}
}
for(int i = 2; i < N; ++i)
{
for(int j = 1; j <= n; ++j)
{
for(int k = 0; k <= 1; ++k)
{
f[i][j][k] = f[i - 1][f[i - 1][j][k]][k];
}
}
}
}
int get_dist(int i, int j)
{
return (int)abs(h[i] - h[j]);
}
void init_d()
{
for(int j = 1; j <= n; ++j)
{
da[0][j][0] = get_dist(j, ga[j]), da[0][j][1] = 0;
db[0][j][1] = 0, db[0][j][1] = get_dist(j, gb[j]);
}
for(int j = 1; j <= n; ++j)
{
for(int k = 0; k <= 1; ++k)
{
da[1][j][k] = da[0][j][k] + da[0][f[0][j][k]][1 - k];
db[1][j][k] = db[0][j][k] + db[0][f[0][j][k]][1 - k];
}
}
for(int i = 2; i < N; ++i)
{
for(int j = 1; j <= n; ++j)
{
for(int k = 0; k <= 1; ++k)
{
da[i][j][k] = da[i - 1][j][k] + da[i - 1][f[i - 1][j][k]][k];
db[i][j][k] = db[i - 1][j][k] + db[i - 1][f[i - 1][j][k]][k];
}
}
}
}
void init()
{
init_g();
init_f();
init_d();
}
void cal(int s, int x, int& la, int& lb)
{
la = lb = 0;
for(int i = N - 1; i >= 0; --i)
{
if(f[i][s][0] && la + lb + da[i][s][0] + db[i][s][0] <= x)
{
la += da[i][s][0];
lb += db[i][s][0];
s = f[i][s][0];
}
}
}
void work()
{
int s, x;
scanf("%d", &x);
int la, lb;
int max_h = 0, ans;
double min_r = inf, r;
for(int i = 1; i <= n; ++i)
{
cal(i, x, la, lb);
r = lb ? 1.0 * la / lb : inf;
if(r < min_r || r == min_r && h[i] > max_h)
{
ans = i;
min_r = r;
max_h = h[i];
}
}
printf("%d\n", ans);
int m;
scanf("%d", &m);
while((m--) > 0)
{
scanf("%d %d", &s, &x);
cal(s, x, la, lb);
printf("%d %d\n", la, lb);
}
}
int main()
{
// freopen("input.txt", "r", stdin);
read();
init();
work();
return 0;
}