BZOJ 2427 & 分块裸题

题意:

　　求区间内的众数,强制在线.

SOL:

　　推荐一个大神犇的blog,讲的还是很好的(主要我喜欢他的代码风格(逃:http://www.cnblogs.com/JoeFan/p/4248767.html

　　太裸没什么意思...虽然好些但码码也挺长的...

　　还是贴那个大神的代码天天看着代码打的会不会好看点>_<

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

using namespace std;

inline void Read(int &Num) {
    char c; c = getchar();
    while (c < '0' || c > '9') c = getchar();
    Num = c - '0'; c = getchar();
    while (c >= '0' && c <= '9') {
        Num = Num * 10 + c - '0';
        c = getchar();
    }
}

const int MaxN = 40000 + 5, MaxBlk = 200 + 5;

int n, m, BlkSize, TotBlk;
int A[MaxN], TL[MaxN], T[MaxN], Cnt[MaxN], L[MaxBlk], R[MaxBlk], First[MaxN], Last[MaxN];
int f[MaxBlk][MaxBlk], g[MaxBlk][MaxBlk];

struct ES
{
    int Pos, Num, v;
} E[MaxN];

inline bool Cmp_Num(ES e1, ES e2) {
    if (e1.Num == e2.Num) return e1.Pos < e2.Pos;
    return e1.Num < e2.Num;
}
inline bool Cmp_Pos(ES e1, ES e2) {return e1.Pos < e2.Pos;}

int GetNum(int Num, int x, int y) {
    if (x > y || x > E[Last[Num]].Pos || y < E[First[Num]].Pos) return 0;
    int l, r, mid, p1, p2;
    l = First[Num]; r = Last[Num];
    while (l <= r) {
        mid = (l + r) >> 1;
        if (E[mid].Pos >= x) {
            p1 = mid;
            r = mid - 1;
        }
        else l = mid + 1;
    }
    l = First[Num]; r = Last[Num];
    while (l <= r) {
        mid = (l + r) >> 1;
        if (E[mid].Pos <= y) {
            p2 = mid;
            l = mid + 1;
        }
        else r = mid - 1;
    }
    return p2 - p1 + 1;
}

int main()
{
    Read(n); Read(m);
    for (int i = 1; i <= n; ++i) {
        Read(E[i].Num);
        E[i].Pos = i;
    }
    sort(E + 1, E + n + 1, Cmp_Num);
    int v_Index = 0;
    for (int i = 1; i <= n; ++i) {//离散化
        if (i == 1 || E[i].Num > E[i - 1].Num) ++v_Index;
        E[i].v = v_Index;
        TL[v_Index] = E[i].Num;//再映射回来
    }
    sort(E + 1, E + n + 1, Cmp_Pos);
    for (int i = 1; i <= n; ++i) A[i] = E[i].v;
    sort(E + 1, E + n + 1, Cmp_Num);
    //三次排序感觉真是浪费啊...
    for (int i = 1; i <= n; ++i) {
        if (First[E[i].v] == 0) First[E[i].v] = i;
        Last[E[i].v] = i;
    }//数字段的头尾
    BlkSize = (int)sqrt((double)n);
    TotBlk = (n - 1) / BlkSize + 1;//很不错的技巧啊
    for (int i = 1; i <= TotBlk; ++i) {//块两头
        L[i] = (i - 1) * BlkSize + 1;
        R[i] = i * BlkSize;
    }
    R[TotBlk] = n;
    for (int i = 1; i <= TotBlk; ++i) {
        for (int j = 1; j <= n; ++j) Cnt[j] = 0;
        f[i][i - 1] = 0; g[i][i - 1] = 0;
        for (int j = i; j <= TotBlk; ++j) {
            f[i][j] = f[i][j - 1];
            g[i][j] = g[i][j - 1];
            for (int k = L[j]; k <= R[j]; ++k) {
                ++Cnt[A[k]];
                if (Cnt[A[k]] > f[i][j] || (Cnt[A[k]] == f[i][j] && A[k] < g[i][j])) {
                    f[i][j] = Cnt[A[k]]; g[i][j] = A[k];//次数和数
                }
            }
        }
    }
    memset(Cnt, 0, sizeof(Cnt));
    for (int i = 1; i <= n; ++i) T[i] = -1;
    int l, r, x, y, Ct, Ans, Cu;
    Ans = 0;
    for (int i = 1; i <= m; ++i) {
        Read(l); Read(r);
        l = (l + Ans - 1) % n + 1; r = (r + Ans - 1) % n + 1;//强制在线
        if (l > r) swap(l, r);
        x = (l - 1) / BlkSize + 1; if (l != L[x]) ++x;
        y = (r - 1) / BlkSize + 1; if (r != R[y]) --y;
        if (x > y) {
            Ct = 0; Ans = 0;
            for (int j = l; j <= r; ++j) {
                ++Cnt[A[j]];
                if (Cnt[A[j]] > Ct || (Cnt[A[j]] == Ct && A[j] < Ans)) {
                    Ct = Cnt[A[j]]; Ans = A[j];
                }
            }
            for (int j = l; j <= r; ++j) --Cnt[A[j]];
        }
        else {
            Ct = f[x][y]; Ans = g[x][y];
            for (int j = l; j < L[x]; ++j) {
                ++Cnt[A[j]];
                if (T[A[j]] == -1) T[A[j]] = GetNum(A[j], L[x], R[y]);
                Cu = Cnt[A[j]] + T[A[j]];
                if (Cu > Ct || (Cu == Ct && A[j] < Ans)) {
                    Ct = Cu; Ans = A[j];
                }
            }
            for (int j = r; j > R[y]; --j) {
                ++Cnt[A[j]];
                if (T[A[j]] == -1) T[A[j]] = GetNum(A[j], L[x], R[y]);
                Cu = Cnt[A[j]] + T[A[j]];
                if (Cu > Ct || (Cu == Ct && A[j] < Ans)) {
                    Ct = Cu; Ans = A[j];
                }
            }
            for (int j = l; j < L[x]; ++j) {--Cnt[A[j]]; T[A[j]] = -1;}
            for (int j = r; j > R[y]; --j) {--Cnt[A[j]]; T[A[j]] = -1;}
        }
        Ans = TL[Ans];
        printf("%d\n", Ans);
    }
    return 0;
}