• hdu 4876 ZCC loves cards(暴力)


    题目链接:hdu 4876 ZCC loves cards

    题目大意:给出n,k,l,表示有n张牌,每张牌有值。选取当中k张排列成圈,然后在该圈上进行游戏,每次选取m(1≤m≤k)张连续的牌,取牌上值的亦或和。要求找到一个圈,使得L~R之间的数都能够得到,输出R。假设R < L输出0.

    解题思路:暴力,首先预处理出来每种选取的亦或值,然后在该基础上从能够组成L的状态中挑选一个,L+1的状态中挑取一个,知道说总的挑取出全部状态中选中的牌的个数大于K为值,然后用全排序去查找最大的R。

    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <map>
    #include <algorithm>
    
    using namespace std;
    typedef long long ll;
    const int maxn = 200;
    const int maxc = 25;
    const int maxs = 1<<20;
    const int maxt = 20005;
    
    bool flag;
    int ans, rec, dp[maxs+5];
    int N, K, L, arr[maxc];
    int vec[maxn][maxt], c[maxn];
    
    inline int bitcount (int s) {
        return s == 0 ? 0 : bitcount(s/2) + (s&1);
    }
    
    void dfs (int d, int s, int val, int f) {
        if (d == K + 1)
            return;
    
        vec[val][c[val]++] = s;
    
        for (int i = f + 1; i < N; i++)
            dfs (d+1, s|(1<<i), val^arr[i], i);
    }
    
    void init () {
        flag = true;
        ans = rec = 0;
        memset(c, 0, sizeof(c));
        memset(dp, 0, sizeof(dp));
    
        for (int i = 0; i < N; i++)
            scanf("%d", &arr[i]);
    
        dfs(0, 0, 0, -1);
    }
    
    int check (int* a) {
        int v[maxn];
        memset(v, 0, sizeof(v));
    
        for (int m = 1; m <= K; m++) {
            int tmp = a[0];
            for (int i = 1; i < m; i++)
                tmp ^= a[i];
    
            for (int i = 0; i < K; i++) {
                v[tmp] = 1;
                tmp ^= (a[i] ^ a[i+m]);
            }
        }
    
        int ans = L;
        while (v[ans] && ans < maxn)
            ans++;
        return ans - 1;
    }
    
    void judge (int s) {
        int cnt = 0, a[2*maxn];
    
        for (int i = 0; i < N; i++) {
            if (s&(1<<i))
                a[cnt++] = arr[i];
        }
    
        sort(a + 1, a + cnt);
        do {
            for (int i = 0; i < cnt; i++)
                a[i+cnt] = a[i];
    
            ans = max(ans, check(a));
        } while (next_permutation(a + 1, a + cnt));
    }
    /*
    */
    
    void solve (int d, int s) {
        int cnt = bitcount(s);
    
        rec = max(d, rec);
        if (cnt >= K || d >= maxn) {
            if (cnt == K) {
                judge(s);
                flag = false;
            }
            return;
        }
    
        for (int i = 0; i < c[d]; i++) {
            int s0 = (s | vec[d][i]);
            if (dp[s0]) continue;
            solve(d+1, s0);
            dp[s0] = 1;
        }
    }
    
    int main () {
        while (scanf("%d%d%d", &N, &K, &L) == 3) {
            init();
            solve(L, 0);
    
            if (flag)
                ans = rec - 1;
            printf("%d
    ", ans < L ? 0 : ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/4008014.html
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