HDU - 5735 Born Slippy 思维 + dp（看题解）

HDU - 5735

感觉这个思路相当巧妙啊。。

考虑最普通的 dp[ i ] = max(dp[ j ] + w[ i ] opt w[ j ])， j 是 i 的祖先。 

把(2 << 16) 分成前八位和后八位去优化最朴素的dp。 修改遍历后八位， 查询遍历前八位。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = (1 << 16) + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = (int)1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, w[N], c[1 << 8];
LL dp[N], f[1 <<  8][1 << 8], g[N][1 << 8];
char op[10];
vector<int> G[N];

inline LL calc(int a, int b) {
    if(*op == 'A') return a & b;
    else if(*op == 'O') return a | b;
    else return a ^ b;
}

void dfs(int u) {
    dp[u] = 0;
    int now_a = w[u] >> 8;
    int now_b = w[u] & 255;
    for(int pre_a = 0; pre_a < 256; pre_a++) {
        if(c[pre_a]) {
            chkmax(dp[u], f[pre_a][now_b] + ((calc(now_a, pre_a)) << 8));
        }
    }
    for(int pre_b = 0; pre_b < 256; pre_b++) {
        g[u][pre_b] = f[now_a][pre_b];
    }
    c[now_a]++;
    for(int pre_b = 0; pre_b < 256; pre_b++) {
        chkmax(f[now_a][pre_b], dp[u] + calc(now_b, pre_b));
    }
    for(auto &v : G[u]) {
        dfs(v);
    }
    c[now_a]--;
    for(int pre_b = 0; pre_b < 256; pre_b++) {
        f[now_a][pre_b] = g[u][pre_b];
    }
}

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%s", &n, op);
        for(int i = 1; i <= n; i++) {
            G[i].clear();
        }
        for(int i = 1; i <= n; i++) {
            scanf("%d", &w[i]);
        }
        for(int i = 2; i <= n; i++) {
            int pa; scanf("%d", &pa);
            G[pa].push_back(i);
        }
        dfs(1);
        int ans = 0;
        for(int i = 1; i <= n; i++) {
            add(ans, 1LL * i * (dp[i] + w[i]) % mod);
        }
        printf("%d\n", ans);
    }
    return 0;
}

/*
*/