This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define pb push_back
#define FIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
using namespace std;
typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
const int N = 307, inf = 2e9;
int n, m, k, siz;
int a[N], b[N];
int mem[2*N][N];
pii gr[2*N][N];
array <int,3> vv[2*N];
int q[6*N], dq[6*N];
vector <int> add[2*N], rem[2*N];
struct MQ {
int type, l1, r1, l2, r2;
inline void push(int x) {q[++r1] = x; while (l2 <= r2 && dq[r2] < x) r2--; dq[++r2] = x;}
inline void pop() {if (dq[l2] == q[l1]) l2++; l1++;}
inline int get() {return dq[l1];}
inline void reset() {l1 = l2 = type*2*N; r1 = r2 = l1-1;}
} m1, m2, m3;
int main () {
FIO;
m1.type = 0;
m2.type = 1;
m3.type = 2;
cin >> n >> m >> k;
vector <pii> so;
for (int i = 0; i < k; i++) {
cin >> a[i] >> b[i];
a[i]--; b[i]--;
so.pb({a[i], b[i]});
}
sort(all(so));
for (int i = 0; i < k; i++) a[i] = so[i].F, b[i] = so[i].S;
int d = a[0]; for (int i = 0; i < k; i++) a[i] -= d;
d = n-a[k-1]-1; for (int i = 1; i < k; i++) d = max(d, a[i]-a[i-1]-1);
a[k] = n-1;
vector <ll> v; v.pb(d);
for (int i = 0; i <= k; i++) {
for (int j = i; j <= k; j++) {
if (a[j]-a[i]-1 >= d) v.pb(a[j]-a[i]-1);
if (a[j]+n-a[i]-1 >= d) v.pb(a[j]+n-a[i]-1);
if (a[i]+n-a[j]-1 >= d) v.pb(a[i]+n-a[j]-1);
}
if (n-a[i]-1 >= d) v.pb(n-a[i]-1);
}
sort(all(v)); v.erase(unique(all(v)), v.end());
//PRECOMPUTE -> O(N^2 log N)
for (int i = 0; i < k; i++) {
set <int> s;
multiset <int> rj;
rj.insert(0);
for (int j = i; j >= 0; j--) {
mem[i][j] = mem[i][j+1];
gr[i][j].F = gr[i][j+1].F;
gr[i][j].S = gr[i][j+1].S;
if (s.find(b[j]) != s.end()) continue;
auto p1 = s.lower_bound(b[j]);
if (p1 == s.end()) {
if (!s.empty()) rj.insert(b[j]-*(--p1)-1);
s.insert(b[j]);
gr[i][j] = {*s.begin(), m-*s.rbegin()-1};
}
else if (p1 == s.begin()) {
rj.insert(*p1-b[j]-1);
s.insert(b[j]);
gr[i][j] = {*s.begin(), m-*s.rbegin()-1};
}
else {
int x = *p1; --p1; int y = *p1;
rj.erase(rj.find(x-y-1));
rj.insert(x-b[j]-1);
rj.insert(b[j]-y-1);
s.insert(b[j]);
}
mem[i][j] = *rj.rbegin();
}
}
//SOLVE O(N^3)
ll mn = inf;
for (auto x : v) {
m1.reset(); m2.reset(); m3.reset(); siz = 0;
int l = 0, r = -1, z = 0;
for (int i = 0; i < k; i++) {
int gt = (x+a[z]+1 < 2LL*n) ? x+a[z]+1 : 2*n;
while (a[z]+x < a[i]) {
if (vv[siz-1][2] != gt) {vv[siz++] = {l+1, r, gt}; l++;}
else vv[siz-1][0]++, l++;
z++;
}
if (!siz || vv[siz-1][2] != a[i]) {vv[siz++] = {l, r+1, a[i]}; r++;}
else vv[siz-1][1]++, r++;
}
int gt = (x+a[z]+1 < 2LL*n) ? x+a[z]+1 : 2*n;
while (z < k) {
if (vv[siz-1][2] != gt) {vv[siz++] = {l+1, r, gt}; l++;}
else vv[siz-1][0]++, l++;
z++;
}
z = 0;
#define C (x+max(m1.get(), m2.get()+m3.get()))
for (int i = 0; i+1 < siz && vv[i][2] < 2*n; i++) {
l = vv[i][0];
r = vv[i][1];
m1.push(mem[r][l]); m2.push(gr[r][l].F); m3.push(gr[r][l].S);
while (vv[z][2]+n-1 < vv[i+1][2]) {
mn = min(mn, C);
m1.pop(); m2.pop(); m3.pop();
z++;
}
}
}
cout << mn << "\n";
return 0;
}
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