[Codeforces 505C] Mr. Kitayuta, the Treasure Hunter
作者:互联网
https://codeforces.ml/contest/505/problem/C
题意:
现有点\(0\)到点\(30000\),每个点都有一个权值。从点\(0\)开始跳,第一步跳\(d\)长。设上一步跳的长度为\(l\),那么下一步能在\([l - 1, l + 1]\)这个区间内选择一个值,以它为步长跳,但是要合法,即步长必须为正整数且不会跳出\(30000\)。
思路:
看着是很裸的\(dp\)或者记忆化搜索的做法,但\(n\)很大,直接\(O(n^2)\)跑过不去。观察一下步长,寻找一下不同的步长最多有多少个。从\(1\)开始,对公差为\(1\)的等差数列求和,发现在第\(245\)项时,和就会超过\(30000\)。但由于我们不知道具体会向哪边扩展,所以我们从\(d\)开始,向两边进行扩展,就能过滤掉永远取不到的步长,得到更优的决策集合,接下来大力推就行了。
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
inline int rd() {
int f = 0; int x = 0; char ch = getchar();
for (; !isdigit(ch); ch = getchar()) f |= (ch == '-');
for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
if (f) x = -x;
return x;
}
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N = 3e4 + 7;
int n, d;
int dp[N][1000], p[N], id[N];
vector<int> G;
void solve() {
n = rd();
d = rd();
for (int i = 1; i <= n; ++i) {
int q = rd();
p[q]++;
}
int tot = 0;
for (int i = max(1, d - 246); i <= min(30000, d + 246); ++i) {
G.push_back(i);
id[i] = tot++;
}
memset(dp, -0x3f, sizeof(dp));
dp[d][ id[d] ] = p[d];
int ans = p[d];
for (int i = d; i <= 30000; ++i) {
for (int j = 0; j < tot; ++j) {
if (dp[i][j] < 0) continue;
for (int k = -1; k <= 1; ++k) {
int to = G[j] + k + i;
if (to <= i || to > 30000) continue;
dp[to][j + k] = max(dp[to][j + k], dp[i][j] + p[to]);
ans = max(ans, dp[to][j + k]);
}
}
}
printf("%d\n", ans);
}
int main() {
int t = 1;
while (t--) solve();
return 0;
}
标签:ch,30000,int,Treasure,Hunter,Codeforces,rd,步长,dp 来源: https://www.cnblogs.com/Sstee1XD/p/14727828.html