Educational Codeforces Round 111 (Rated for Div. 2) E. Stringforces 二分答案+状压dp

Posted 2021-08-20 kaka0010

原题链接：https://codeforces.ml/contest/1550/problem/E

目录

• 题意
• 分析
• Code

题意

有一个字符串由k个字符组成’?'可以填任意字符，字符有一个权值为$fi$表示最长连续相同子串的长度，问所有字符中最小的$fi$最大可以是多少。

分析

看到$Max(\sum Min$)问题，最自然想到二分答案，先确定一个长度，然后去填数，判断是否可以满足。

首先初步的构想就是这样，然后考虑优化。如果直接暴力去填数，每个位置都要枚举k种情况，也就是$2^{k}nlogn$的复杂度，那么我们想如果可以直接预处理出某个位置完成填数的最近距离，然后再用状压DP去跑是不是就可以了。

于是先预处理一个数组$f[i][j]$代表第i个字符从第j个位置开始，最先完成填数的位置，这个我们倒着跑一遍就可以了。然后状压处理起来就简单了，我们用dp[i]表示处理完当前字符的最近位置，转移方程就是
$$dp[i|(1<<j)]=min(dp[i|(1<<j)],f[j][dp[i])$$

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const ll INF = 1e18;
const int N = 2e5 + 10;
const int M = 1e6 + 10;
const int MOD = 1e9 + 7;
char s[N];
int dp[1<<17];
int f[20][N], n, k;
bool check(int x) {
    memset(dp, 0x3f, sizeof dp);
    memset(f, 0x3f, sizeof f);
    for (int i = 0; i < k; i++) {
        int now = 0;
        for (int j = n; j >= 1; j--) {
            if (s[j] == '?' || s[j] == 'a' + i) now++;
            else now = 0;
            if (now >= x) f[i][j] = j + x;
            else f[i][j] = f[i][j+1];
        }
    }
    dp[0] = 1;
    for (int i = 0; i < 1 << k; i++) {
        for (int j = 0; j < k; j++) {
            if (i >> j & 1) continue;
            int state = i | (1 << j);
            if (dp[i] <= n + 1) dp[state] = min(dp[state], f[j][dp[i]]);
        }
    }
    if (dp[(1<<k)-1] > n + 1) return false;
    else return true;
}
void solve() {
    cin >> n >> k;
    cin >> (s + 1);
    int l = 0, r = n;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid)) l = mid + 1;
        else r = mid - 1;
    }
    cout << r << endl;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
    signed test_index_for_debug = 1;
    char acm_local_for_debug = 0;
    do {
        if (acm_local_for_debug == '$') exit(0);
        if (test_index_for_debug > 20)
            throw runtime_error("Check the stdin!!!");
        auto start_clock_for_debug = clock();
        solve();
        auto end_clock_for_debug = clock();
        cout << "Test " << test_index_for_debug << " successful" << endl;
        cerr << "Test " << test_index_for_debug++ << " Run Time: "
             << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
        cout << "--------------------------------------------------" << endl;
    } while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#else
    solve();
#endif
    return 0;
}