HDU5845 Best Division

递归写法，好久不写很容易就gg了...

 

dp[i]=max(dp[j])+1,并且s[i]XORs[j]<=x 

01字典树优化一下转移。

  1 #include <bits/stdc++.h>
  2 
  3 #define ll long long
  4 #define ull unsigned long long
  5 #define st first
  6 #define nd second
  7 #define pii pair<int, int>
  8 #define pil pair<int, ll>
  9 #define pli pair<ll, int>
 10 #define pll pair<ll, ll>
 11 #define tiii tuple<int, int, int>
 12 #define pw(x) ((1LL)<<(x))
 13 #define lson l, m, rt<<1
 14 #define rson m+1, r, rt<<1|1
 15 #define FIN freopen("A.in","r",stdin);
 16 #define FOUT freopen("A.out","w",stdout);
 17 using namespace std;
 18 /***********/
 19 template <class T>
 20 bool scan (T &ret) {
 21     char c;
 22     int sgn;
 23     if (c = getchar(), c == EOF) return 0; //EOF
 24     while (c != '-' && (c < '0' || c > '9') ) c = getchar();
 25     sgn = (c == '-') ? -1 : 1;
 26     ret = (c == '-') ? 0 : (c - '0');
 27     while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
 28     ret *= sgn;
 29     return 1;
 30 }
 31 template<typename N,typename PN>inline N flo(N a,PN b){return a>=0?a/b:-((-a-1)/b)-1;}
 32 template<typename N,typename PN>inline N cei(N a,PN b){return a>0?(a-1)/b+1:-(-a/b);}
 33 template<typename T>inline int sgn(T a) {return a>0?1:(a<0?-1:0);}
 34 template <class T1, class T2>
 35 bool gmax(T1 &a, const T2 &b) { return a < b? a = b, 1:0;}
 36 template <class T1, class T2>
 37 bool gmin(T1 &a, const T2 &b) { return a > b? a = b, 1:0;}
 38 template <class T> inline T lowbit(T x) {return x&(-x);}
 39 
 40 template<class T1, class T2>
 41 ostream& operator <<(ostream &out, pair<T1, T2> p) {
 42     return out << "(" << p.st << ", " << p.nd << ")";
 43 }
 44 template<class A, class B, class C>
 45 ostream& operator <<(ostream &out, tuple<A, B, C> t) {
 46     return out << "(" << get<0>(t) << ", " << get<1>(t) << ", " << get<2>(t) << ")";
 47 }
 48 template<class T>
 49 ostream& operator <<(ostream &out, vector<T> vec) {
 50     out << "("; for(auto &x: vec) out << x << ", "; return out << ")";
 51 }
 52 void testTle(int &a){
 53     while(1) a = a*(long long)a%1000000007;
 54 }
 55 const int inf = 0x3f3f3f3f;
 56 const long long INF = 1e17;
 57 const long long mod = 1000000007;
 58 const double eps = 1e-6;
 59 const int N = 1e5+5;
 60 /***********/
 61 
 62 int a[N], X;
 63 struct Tire{
 64     int tot, node[N][2], val[N], dp[N];
 65     int newnode(){
 66         ++tot;
 67         node[tot][0] = node[tot][1] = val[tot] = 0;
 68         dp[tot] = -1e9;
 69         return tot;
 70     }
 71     void init(){
 72         tot = 0;
 73         newnode();
 74     }
 75     void pushup(int rt){
 76         val[rt] = 0;
 77         dp[rt] = -1e9;
 78         if(node[rt][0]&&val[ node[rt][0] ]) {
 79             gmax(dp[rt], dp[ node[rt][0] ]);
 80             val[rt] += val[ node[rt][0] ];
 81         }
 82         if(node[rt][1]&&val[ node[rt][1] ]) {
 83             gmax(dp[rt], dp[ node[rt][1] ]);
 84             val[rt] += val[ node[rt][1] ];
 85         }
 86     }
 87     void update(int x, int d, int v, int now = 1, int dep = 30){
 88         if(dep == -1){
 89             val[now] += d;
 90             if(val[now]) gmax(dp[now], v);
 91             else dp[now] = -1e9;
 92             return ;
 93         }
 94         int t = 1&(x>>dep);
 95         if(!node[now][t]) node[now][t] = newnode();
 96         update(x, d, v, node[now][t], dep-1);
 97         pushup(now);
 98     }
 99     int query(int x, int now = 1, int dep = 30){
100         if(dep == -1) return dp[now];
101         int t1 = 1&(X>>dep), t2 = 1&(x>>dep);
102         int ret = -1e9;
103         if(t1){
104             if(node[now][t2]) gmax(ret, dp[ node[now][t2] ]);
105             if(node[now][t2^1]) gmax(ret, query(x, node[now][t2^1], dep-1));
106         }
107         else if(node[now][t2]) gmax(ret, query(x, node[now][t2], dep-1));
108         return ret;
109     }
110 }T;
111 int f[N];
112 int main(){
113     int t; scanf("%d", &t);
114     while(t--){
115         int n, L, p, q, i;
116         scanf("%d%d%d", &n, &X, &L);
117         scanf("%d%d%d", a+1, &p, &q);
118         for(i = 2; i <= n; i++)
119             a[i] = (a[i-1]*(ll)p+q)%268435456LL;
120         for(int i = 2; i <= n; i++)
121             a[i] = a[i]^a[i-1];
122         T.init();
123         T.update(a[0], 1, f[0]);
124         for(i = 1; i <= n&&i <= L; i++){
125             f[i] = T.query(a[i])+1;
126             T.update(a[i], 1, f[i]);
127         }
128 
129         for( ; i <= n; i++){
130             T.update(a[i-L-1], -1, f[i-L-1]);
131             f[i] = T.query(a[i])+1;
132             T.update(a[i], 1, f[i]);
133         }
134         printf("%d\n", max(f[n], 0));
135     }
136     return 0;
137 }
View Code

 

posted @ 2016-09-08 17:02  我在地狱  阅读(366)  评论(1)  编辑  收藏  举报