hdu 5715 XOR 游戏(trie)

看这里

dp[i][k] = max(min(dp[j][k-1], s[j] ^ s[i])，dp[i][k])。

二分答案val，则dp[j][k-1]肯定要满足大于等于val，并且s[j] ^ s[i] >= val才能够转移。

在trie树上找最大的 s[j] ^ s[i]。。

#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define pii pair<int, int>
#define MP make_pair
#define ls i << 1
#define rs ls | 1
#define md (ll + rr >> 1)
#define lson ll, md, ls
#define rson md + 1, rr, rs
#define inf 0x3f3f3f3f
#define mod 1000000007
#define eps 1e-8
#define N 10010
#define M 200020

struct trie{
int son[N*32][2], tim[N*32], tot;
void init(){
son[0][0] = son[0][1] = tot = 0;
tim[0] = 1;
}
void insert(int val){
int t = 0;
for(int i = 30; i >= 0; --i){
int c = (val >> i) & 1;
if(son[t][c] == 0){
son[t][c] = ++tot;
son[tot][0] = son[tot][1] = tim[tot] = 0;
}
t = son[t][c];
tim[t]++;
}
}
void del(int val){
int t = 0;
for(int i = 30; i >= 0; --i){
int c = (val >> i) & 1;
int tt = t;
t = son[t][c];
tim[t]--;
if(tim[t] == 0){
son[tt][c] = 0;
break;
}
}
}
int query(int val){
int ret = 0, t = 0;
for(int i = 30; i >= 0; --i){
int c = (val >> i) & 1;
if(son[t][c^1]){
t = son[t][c^1];
ret += (1 << i);
}
else {
t = son[t][c];
}
}
return ret;
}
}tree[15];

int n, m, K, ans, a
, s
;
int dp
[15];
bool check(int val){
dp[0][0] = 1;
for(int i = 0; i <= m; ++i) tree[i].init();
tree[0].insert(0);
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= m; ++j){
dp[i][j] = 0;
if(i > K && dp[i- K - 1][j-1])
tree[j-1].del(s[i- K - 1]);
}
for(int j = m; j >= 1; --j){
if(tree[j-1].query(s[i]) >= val){
dp[i][j] = 1;
tree[j].insert(s[i]);
}
}
}
return dp
[m];
}
int main(){
int cas, kk = 0;
scanf("%d", &cas);
while(cas--){
scanf("%d%d%d", &n, &m, &K);
for(int i = 1; i <= n; ++i)
scanf("%d", &a[i]), s[i] = s[i-1] ^ a[i];

int L = 0, R = 1000000010;
while(L < R){
int mid = (L + R) >> 1;
if(check(mid))
ans = mid, L = mid + 1;
else R = mid;
}
printf("Case #%d:\n%d\n", ++kk, ans);
}
return 0;
}