# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
373152 | LucaDantas | Collecting Stamps 3 (JOI20_ho_t3) | C++17 | 43 ms | 72812 KiB |
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>
#define ff first
#define ss second
constexpr int maxn = 200+10;
std::pair<int,int> a[maxn];
int dp[maxn][maxn][maxn][2];
void min_self(int& a, int b) { a = std::min(a, b); }
int main() {
int n, T; scanf("%d %d", &n, &T);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i].ff);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i].ss);
a[0] = {0, -1}; // a origem
++n;
auto prev = [&](int x) {
return (x+n-1)%n;
};
auto next = [&](int x) {
return (x+1)%n;
};
auto dist = [&](int x, int y) {
if(x > y) return y + T - x;
return y - x;
};
memset(dp, 0x3f, sizeof dp);
dp[0][0][0][0] = 0;
for(int sz = 0; sz < n; sz++) {
for(int l = 0; l < n; l++) {
int r = (l+sz)%n;
for(int k = 0; k <= sz; k++) {
if(std::min(dp[l][r][k][0], dp[l][r][k][1]) == 0x3f3f3f3f || l == r+1) continue;
// printf("%d %d %d (%d %d)\n\n", l, r, k, dp[l][r][k][0], dp[l][r][k][1]);
// eu ligando com o anterior com o ptr na esq do itv
int t = dp[l][r][k][0] + dist(a[prev(l)].first, a[l].first);
min_self(dp[prev(l)][r][k+(t<=a[prev(l)].second)][0], t);
// printf("%d %d\n", t, prev(l));
// eu ligando com o anterior com o ptr na dir do itv
t = dp[l][r][k][1] + dist(a[prev(l)].first, a[r].first);
min_self(dp[prev(l)][r][k+(t<=a[prev(l)].second)][0], t);
// eu ligando com o proximo com o ptr na esq do itv
t = dp[l][r][k][0] + dist(a[l].first, a[next(r)].first);
min_self(dp[l][next(r)][k+(t<=a[next(r)].second)][1], t);
// eu ligando com o proximo com o ptr na dir do itv
t = dp[l][r][k][1] + dist(a[r].first, a[next(r)].first);
min_self(dp[l][next(r)][k+(t<=a[next(r)].second)][1], t);
}
}
}
int ans = 0;
for(int k = 0; k <= n; k++)
for(int l = 0; l <= n; l++)
for(int r = 0; r <= n; r++)
for(int b : {0, 1})
if(dp[l][r][k][b] < 0x3f3f3f3f)
ans = k;
printf("%d\n", ans);
}
Compilation message (stderr)
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