2015北京网络赛 J Scores bitset+分块

2015北京网络赛 J Scores

题意：50000组5维数据，50000个询问，问有多少组每一维都不大于询问的数据

思路：赛时没有思路，后来看解题报告也因为智商太低看了半天看不懂。bitset之前没用过，查了下发现其实就是一个二进制表示，这里的每一位就表示原序中的状态。

　　  建一个bitset<50000> bs[6][sqrt(50000)], bs[i][j] 就表示在第i维上前j个块中的数据在原序中是哪些位置。

 #include <iostream>
 #include <cstdio>
 #include <fstream>
 #include <algorithm>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <string>
 #include <cstring>
 #include <map>
 #include <stack>
 #include <set>
 #include <bitset>
 #define LL long long
 #define eps 1e-8
 #define INF 0x3f3f3f3f
 #define MAXN 50005
 using namespace std;
 struct Node{
     int x, pos;
 };
 bitset<MAXN> bs[][];
 int t[];
 bitset<MAXN> res[], w;
 bool compare(Node a, Node b){
     return a.x < b.x;
 }
 Node s[][MAXN];
 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     //freopen("out.txt", "w", stdout);
 #endif // OPEN_FILE
     int T;
     int n, m;
     scanf("%d", &T);
     while (T--){
         scanf("%d%d", &n, &m);
         for (int i = ; i <= n; i++){
             for (int j = ; j <= ; j++){
                 scanf("%d", &s[j][i].x);
                 s[j][i].pos = i;
             }
         }
         for (int i = ; i <= ; i++){
             sort(s[i] + , s[i] +  + n, compare);
         }
         int u = sqrt(n);
         int num = n / u;
         if (n % u != ){
             num++;
         }
         memset(bs, , sizeof(bs));
         for (int i = ; i <= ; i++){
             for (int j = ;j <= num; j++){
                 bs[i][j] |= bs[i][j - ];
                 int left = (j - ) * u + , right = j * u;
                 if (right > n){
                     right = n;
                 }
                 for (int k = left; k <= right; k++){
                     bs[i][j][s[i][k].pos] = ;
                 }
             }
         }
         int ans = ;
         int q;
         scanf("%d", &q);
         while (q--){
             for (int i = ; i <= ; i++){
                 scanf("%d", &t[i]);
                 t[i] ^= ans;
             }
             for (int i = ; i <= ; i++){
                 int left = , right = n;
                 while (left < right){
                     int mid = (left + right) >> ;
                     if (s[i][mid].x > t[i]){
                         right = mid;
                     }
                     else{
                         left = mid + ;
                     }
                 }
                 int pos;
                 if (s[i][left].x > t[i]){
                     pos = left - ;
                 }
                 else{
                     pos = left;
                 }
                 int u_pos = (pos - ) / u;
                 res[i] = bs[i][u_pos];
                 for (int j = u_pos * u + ; j <= pos; j++){
                     res[i][s[i][j].pos] = ;
                 }
             }
             w = res[];
             for (int i = ; i <= ; i++){
                 w &= res[i];
             }
             ans = w.count();
             printf("%d\n", ans);
         }
     }
 }