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[6][230];
int t[6];
bitset<MAXN> res[6], w;
bool compare(Node a, Node b){
    return a.x < b.x;
}
Node s[6][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 = 1; i <= n; i++){
            for (int j = 1; j <= 5; j++){
                scanf("%d", &s[j][i].x);
                s[j][i].pos = i;
            }
        }
        for (int i = 1; i <= 5; i++){
            sort(s[i] + 1, s[i] + 1 + n, compare);
        }
        int u = sqrt(n);
        int num = n / u;
        if (n % u != 0){
            num++;
        }
        memset(bs, 0, sizeof(bs));
        for (int i = 1; i <= 5; i++){
            for (int j = 1;j <= num; j++){
                bs[i][j] |= bs[i][j - 1];
                int left = (j - 1) * u + 1, right = j * u;
                if (right > n){
                    right = n;
                }
                for (int k = left; k <= right; k++){
                    bs[i][j][s[i][k].pos] = 1;
                }
            }
        }
        int ans = 0;
        int q;
        scanf("%d", &q);
        while (q--){
            for (int i = 1; i <= 5; i++){
                scanf("%d", &t[i]);
                t[i] ^= ans;
            }
            for (int i = 1; i <= 5; i++){
                int left = 1, right = n;
                while (left < right){
                    int mid = (left + right) >> 1;
                    if (s[i][mid].x > t[i]){
                        right = mid;
                    }
                    else{
                        left = mid + 1;
                    }
                }
                int pos;
                if (s[i][left].x > t[i]){
                    pos = left - 1;
                }
                else{
                    pos = left;
                }
                int u_pos = (pos - 1) / u;
                res[i] = bs[i][u_pos];
                for (int j = u_pos * u + 1; j <= pos; j++){
                    res[i][s[i][j].pos] = 1;
                }
            }
            w = res[1];
            for (int i = 2; i <= 5; i++){
                w &= res[i];
            }
            ans = w.count();
            printf("%d
", ans);
        }
    }
}