洛谷 P4168 蒲公英 分块入门

我学分块入门的第二题，其实它应该作为第一题做的

虽然之前做过一道类似的题目：作诗， 但是还是调试了很长的时间， 很多细节的地方经常搞错

看来代码还是得多练

 

先记录一下代码：

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 40010;
const int maxm = 210;

struct u {
    int v,i;
    bool operator < (const u &rhs) const {
      return v < rhs.v;
    }
}B[maxn];

int n, m, N, S, c;
int A[maxn], F[maxm][maxm], G[maxm][maxn], ID[maxn], cnt[maxn], cntM[maxn];
int rk[maxn], col[maxn], frk[maxn];

bool cmp(u x,u y) {
    return x.i < y.i;
}

void lsh() {
    sort(B+1,B+n+1);
    for(int i = 1;i <= n;i++) {
        if(B[i].v != col[B[i-1].v]) c++;
        col[c] = B[i].v;
        B[i].v = c;
    }
    sort(B+1,B+n+1,cmp);
    for(int i = 1;i <= n;i++) A[i] = B[i].v;
}

int query(int x,int y,int t) {
    int ret = 0, MX, MX_i;
    
    if(ID[x] + 1 >= ID[y]) {
      MX = 0, MX_i = 0;
        for(int i = x;i <= y;i++) {
            if(cntM[A[i]] != t) {
                cntM[A[i]] = t;
                cnt[A[i]] = 0;
            }
            cnt[A[i]]++;
            int tp = cnt[A[i]];
            if(tp > MX || (tp == MX && A[i] < MX_i)) {
                MX = tp;
                MX_i = A[i];
            }
        }
        return col[MX_i];
    }
    
    ret = F[ID[x]+1][ID[y]-1];
    MX = G[ID[y]-1][ret] - G[ID[x]][ret], MX_i = ret;
    for(int i = x;i <= y;) {
        if(cntM[A[i]] != t) {
            cntM[A[i]] = t;
            cnt[A[i]] = 0;
        }
    
        cnt[A[i]]++;
        int tp = cnt[A[i]] + G[ID[y]-1][A[i]] - G[ID[x]][A[i]];
        //cout<<ID[y-1]*S+1<<endl; 
        if(tp > MX || (tp == MX && A[i] < MX_i)) {
            MX = tp;
            MX_i = A[i];
        }
        
        if(i == ID[x]*S) i = (ID[y]-1)*S+1;
        else i++;
    }
    
    return col[MX_i];
}

int main() {
    //freopen("d.txt","r",stdin);
    scanf("%d %d",&n,&m);
    S = sqrt(n);
    N = (n-1)/S + 1;
    for(int i = 1;i <= n;i++) {
        scanf("%d",&A[i]);
        B[i].v = A[i];
        B[i].i = i;
        ID[i] = (i-1)/S + 1;
    }
    
    lsh();
    
    int Max = 0, Max_i = n+1;
    for(int i = 1;i <= N;i++) {
        int t = i;
        Max = 0;
        Max_i = 0;
        for(int j = S*(i-1)+1; j <= n;j++) {
            int p = ID[j];
            if(cntM[A[j]] != t) {
                cnt[A[j]] = 0;
                cntM[A[j]] = t;
            }
            cnt[A[j]]++;
            int tp = cnt[A[j]];
            if(tp > Max || (tp == Max && Max_i > A[j])) {
                Max = tp;
                Max_i = A[j];
            }
            if(j%S == 0) F[i][p] = Max_i;
        }
    }

    memset(cnt,0,sizeof(cnt));
    for(int i = 1;i <= n;i++) {
        cnt[A[i]]++;
        if(i%S == 0 || i == n) {
            for(int j = 1;j <= c;j++) {
                G[ID[i]][j] = cnt[j];
            }
        }
    }

    int ans = 0;
    for(int i = 1,x,y;i <= m;i++) {
        scanf("%d %d",&x,&y);
        x = (x+ans-1)%n + 1;
        y = (y+ans-1)%n + 1;
        if(x > y) swap(x,y);
    //    cout <<x<< " : "<<y<<endl;
        ans = query(x,y,i+n);
      printf("%d\n", ans);
    }
    
    return 0;
}