CodeForces 1841C Ranom Numbers

洛谷传送门

CF 传送门

先反转 \(s\) 串，然后考虑 dp，设 \(f_{i, j, 0/1}\) 为考虑了 \([1, i]\)，前缀最大值为 \(j\)，是否修改过字符的最大得分。

转移讨论是否在这个位置修改即可。

时间复杂度 \(O(n)\)。

code

// Problem: C. Ranom Numbers
// Contest: Codeforces - Educational Codeforces Round 150 (Rated for Div. 2)
// URL: https://codeforces.com/group/bXwyZVR0kC/contest/1841/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 200100;
const int mp[] = {1, 10, 100, 1000, 10000};

int n, a[maxn];
ll f[maxn][5][2];
char s[maxn];

void solve() {
	scanf("%s", s + 1);
	n = strlen(s + 1);
	reverse(s + 1, s + n + 1);
	for (int i = 1; i <= n; ++i) {
		a[i] = s[i] - 'A';
	}
	for (int i = 0; i <= n; ++i) {
		for (int j = 0; j < 5; ++j) {
			for (int k = 0; k < 2; ++k) {
				f[i][j][k] = -1e18;
			}
		}
	}
	f[0][0][0] = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j < 5; ++j) {
			for (int k = 0; k < 2; ++k) {
				if (f[i - 1][j][k] < -1e17) {
					continue;
				}
				if (j > a[i]) {
					f[i][j][k] = max(f[i][j][k], f[i - 1][j][k] - mp[a[i]]);
				} else {
					f[i][a[i]][k] = max(f[i][a[i]][k], f[i - 1][j][k] + mp[a[i]]);
				}
			}
			for (int k = 0; k < 5; ++k) {
				if (f[i - 1][j][0] < -1e17) {
					continue;
				}
				if (j > k) {
					f[i][j][1] = max(f[i][j][1], f[i - 1][j][0] - mp[k]);
				} else {
					f[i][k][1] = max(f[i][k][1], f[i - 1][j][0] + mp[k]);
				}
			}
		}
	}
	ll ans = -1e18;
	for (int i = 0; i < 5; ++i) {
		for (int j = 0; j < 2; ++j) {
			ans = max(ans, f[n][i][j]);
		}
	}
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}