BZOJ - 3578: GTY的人类基因组计划2

3578: GTY的人类基因组计划2

第一次居然hash被卡了。改了改rd()就A了。

题解：咱给每个人都随机一个数，几个人就是把他们的数异或起来，用set判重。再开一个set记录合法的房间。

#include<set>
#include<cstdio>
#include<bitset>
#include<cstdlib>
#include<iostream>
using namespace std;
inline char nc() {
    static char b[1<<16],*s=b,*t=b;
    return s==t&&(t=(s=b)+fread(b,1,1<<16,stdin),s==t)?-1:*s++;
}
inline void read(int &x) {
    char b = nc(); x = 0;
    for (; !isdigit(b); b = nc());
    for (; isdigit(b); b = nc()) x = x * 10 + b - '0';
}
inline int op() {
    char b = nc(); for (; b != 'C' && b != 'W'; b = nc()); return b == 'C';
}
typedef unsigned long long ull;
typedef ull bt;
int n, m, q, p[100005], c[100005];
const ull ULL_MAX = -1;
bt a[100005], b[100005];
inline ull rd() {
    return ull(rand()) * rand() * rand();
}
set < bt > h;
set < int > s;
int main() {
    read(n); read(m); read(q); srand(89513255); s.insert(1);
    for (int i = 1; i <= n; ++i) a[i] = rd(), p[i] = 1;
    for (int i = 1; i <= n; ++i) b[1] ^= a[i]; c[1] = n;
    for (int l, r, o, i = 0; i < q; ++i) {
        o = op(); read(l); read(r);
        if (o) {
            if (p[l] == r) continue;
            s.erase(p[l]); s.erase(r);
            b[p[l]] ^= a[l]; --c[p[l]];
            if (!h.count(b[p[l]])) s.insert(p[l]);
            b[r] ^= a[l]; ++c[r]; p[l] = r;
            if (!h.count(b[r])) s.insert(r);
        } else {
            int ans = 0; static set < int > :: iterator it; it = s.lower_bound(l);
            for (; it != s.end() && *it <= r; it = s.lower_bound(l)) {
                h.insert(b[*it]); ans += c[*it]; s.erase(it);
            } printf("%d\n", ans);
        }
    }
    return 0;    
}