Foj 2296 Alice and Bob

Foj 2296 Alice and Bob

题意

题解

代码

扩展

game dfs 写法需考虑的状态表示问题需要多练习。
dp 需练习。

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define sz(a) (int)a.size()
#define de(a) cout << #a << " = " << a << endl
#define dd(a) cout << #a << " = " << a << " "
#define all(a) a.begin(), a.end()
#define endl "\n"
typedef long long ll;
typedef pair<int, int> pii;
//---

int n;
int dp[11][11][11], cnt[10];

int dfs(int a, int b, int c, int d, int pre, int cnt1, int cnt2, int ty) {
    if(a+b+c+d == 0) return pre==0 ? cnt1 : cnt2;
    if(ty == 0) {
        int ans = 0;
        if(a) ans = max(ans, dfs(a-1, b, c, d, pre, cnt1+1, cnt2, 1));
        if(b) ans = max(ans, dfs(a, b-1, c, d, pre^1, cnt1+1, cnt2, 1));
        if(c) ans = max(ans, dfs(a, b, c-1, d, pre, cnt1, cnt2, 1));
        if(d) ans = max(ans, dfs(a, b, c, d-1, pre^1, cnt1, cnt2, 1));
        return ans;
    } else {
        int ans = 10;
        if(a) ans = min(ans, dfs(a-1, b, c, d, pre, cnt1, cnt2+1, 0));
        if(b) ans = min(ans, dfs(a, b-1, c, d, pre, cnt1, cnt2+1, 0));
        if(c) ans = min(ans, dfs(a, b, c-1, d, pre, cnt1, cnt2, 0));
        if(d) ans = min(ans, dfs(a, b, c, d-1, pre, cnt1, cnt2, 0));
        return ans;
    }
}

void init() {
    rep(i, 0, 11) rep(j, 0, 11-i) rep(k, 0, 11-i-j) dp[i][j][k] = dfs(i, j, k, 10-i-j-k, 0, 0, 0, 0);
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    init();
    int T;
    cin >> T;
    while(T--) {
        ///
        cin >> n;
        ///init
        memset(cnt, 0, sizeof(cnt));
        ///solve
        int k = n%10;
        int ans = n = n/10;
        while(n) {
            ++cnt[n%10];
            n/=10;
        }
        int a = 0, b = 0, c = 0;
        rep(i, 0, k) {
            if(cnt[i]%2==0) ++a;
            else ++b;
        }
        rep(i, k, 10) c += cnt[i]%2==0;
        cout << ans*5 + dp[a][b][c] << endl;
    }
    return 0;
}