kyeremal-bzoj2038-[2009国家集训队]-小z的袜子(hose)-莫队算法

bzoj2038-[2009国家集训队]-小z的袜子(hose)

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Description

Input

Output

Sample Input

Sample Output

HINT

Source

大意:给出一堆数,每次询问一个区间,求区间内任取两个数相同的概率.

分析:orz莫涛,莫队专治各种区间询问

给出几篇关于莫队算法的博客:

很通俗直观的莫对算法理解

详细的本题题解

严谨的平面点曼哈顿最小距离生成树论文

更加严谨的英文论文

code:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>

using namespace std;

#define rep(i, l, r) for (LL i = (l); (i) <= (r); ((i)++))
#define REP(i, l, r) for (LL i = (l); (i) >= (r); ((i)--))
#define LL long long
#define MAXN 1000010

LL n, m, col[MAXN], len[MAXN], K, c[MAXN], pos[MAXN], fz, fm, minc, maxc;
struct qr {LL l, r, num;} a[MAXN];
struct answer {LL x, y, num;} ans[MAXN];

inline LL min(LL a, LL b) {return a<b ? a : b;}
inline LL max(LL a, LL b) {return a>b ? a : b;}
inline bool cmp(qr a, qr b) {return pos[a.l]<pos[b.l] || (pos[a.l]==pos[b.l] && a.r<b.r);}
inline LL com(LL n) {return (n-1)*n/2;}
inline answer mp(LL x, LL y, LL num) {answer t; t.x = x, t.y = y, t.num = num; return t;}
inline bool cmp2(answer a, answer b) {return a.num < b.num;}
inline LL gcd(LL a, LL b) {return !b ? a : gcd(b, a%b);}

inline void update(LL &L, LL &R, LL l, LL r, LL &fz, LL &fm) {// [L, R] -> [l, r]
    if (l < L) REP(i, L-1, l) fz += c[col[i]], fm += R-L+1, c[col[i]]++, L--;
    else if (l > L) rep(i, L, l-1) fz += -c[col[i]] + 1, fm += L-R, c[col[i]]--, L++;
    if (r > R) rep(i, R+1, r) fz += c[col[i]], fm += R-L+1, c[col[i]]++, R++;
    else if (r < R) REP(i, R, r+1) fz += -c[col[i]] + 1, fm += L-R, c[col[i]]--, R--;
}

inline void block() {
    K = sqrt(n);
    rep(i, 1, K-1) len[i] = K;
    len[K] = n - (K-1)*K;
    rep(i, 1, K) rep(j, (i-1)*K + 1, (i-1)*K + len[i]) pos[j] = i;
    sort(a+1, a+1+m, cmp);
}

inline void xx() {
    memset(c, 0, sizeof(c));
    LL L = a[1].l, R = a[1].r;
    rep(i, L, R) c[col[i]]++;
    rep(i, minc, maxc) fz += (c[i] * (c[i]-1) / 2);
    fm = com(R-L+1);
    ans[1] = mp(fz, fm, a[1].num);
    rep(i, 2, m) update(L, R, a[i].l, a[i].r, fz, fm), ans[i] = mp(fz, fm, a[i].num);
}

int main() {
    minc = MAXN, maxc = 0;
    cin >> n >> m;
    rep(i, 1, n) scanf("%lld", col + i), maxc = max(maxc, col[i]), minc = min(minc, col[i]);
    rep(i, 1, m) scanf("%lld%lld", &a[i].l, &a[i].r), a[i].num = i;
    block();
    xx();
    sort(ans+1, ans+1+m, cmp2);
    rep(i, 1, m) n = gcd(ans[i].x, ans[i].y), printf("%lld/%lld\n", ans[i].x/n, ans[i].y/n);
    
    return 0;
}