分块

I Hate It

HDU - 1754

 

 

//fen kuai
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
const int inf = 0x3f3f3f3f;
int n, m, SZ;
int a[maxn], bc[maxn], Max[maxn];

void update(int p, int v){
    int tp = (p - 1) / SZ + 1;
    a[p] = v;
    int temp = -inf;
    for(int i = (tp - 1) * SZ + 1; i <= tp * SZ; i++){
        temp = max(temp, a[i]);
    }
    Max[tp] = temp;
}

int query(int l, int r){
    int ans = -inf;
    int pl = (l - 1) / SZ + 1;
    for(int i = l; i <= min(r, pl * SZ); i++){
        ans = max(ans, a[i]);
    }
    int pr = (r - 1) / SZ + 1;
    if(pl != pr){
        for(int i = (pr - 1) * SZ + 1; i <= r; i++){
            ans = max(ans, a[i]);
        }
    }
    for(int i = pl + 1; i < pr; i++){
        ans = max(ans, Max[i]);
    }
    return ans;
}

int main(){
    while(scanf("%d %d", &n, &m) != EOF){
        memset(Max, -inf, sizeof Max);
        SZ = sqrt(n * 1.0);
        for(int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
            bc[i] = (i - 1) / SZ + 1;
            Max[bc[i]] = max(Max[bc[i]], a[i]);
        }
        while(m--){
            char op[4];
            int l, r;
            scanf("%s %d %d", op, &l, &r);
            if(op[0] == 'Q') {
                printf("%d
", query(l, r));
            }else{
                update(l, r);
            }
        }
    }
    return 0;
}

A Simple Problem with Integers

POJ - 3468

//fen kuai
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
using namespace std;
#define LL long long
const int maxn = 100010;
LL a[maxn], atag[maxn];
LL sum[maxn];
int n, m, SZ;

LL query(int l, int r){
    LL ans = 0;
    int pl = (l - 1) / SZ + 1;
    int pr = (r - 1) / SZ + 1;
    for(int i = l; i <= min(r, pl * SZ); i++){
        ans += a[i] + atag[pl];
    }
    if(pl != pr){
        for(int i = (pr - 1) * SZ + 1; i <= r; i++){
            ans += a[i] + atag[pr];
        }
    }
    for(int i = pl + 1; i < pr; i++){
        ans += sum[i];
    }
    return ans;
}

void update(int l, int r, LL c){
    int pl = (l - 1) / SZ + 1;
    int pr = (r - 1) / SZ + 1;
    for(int i = l; i <= min(r, pl * SZ); i++){
        a[i] += c;
        sum[pl] += c;
    }
    if(pl != pr){
        for(int i = (pr - 1) * SZ + 1; i <= r; i++){
            a[i] += c;
            sum[pr] += c;
        }
    }
    for(int i = pl + 1; i < pr; i++){
        sum[i] += c * SZ;
        atag[i] += c;
    }
}

int main(){
    while(scanf("%d %d", &n, &m) != EOF){
        SZ = sqrt(n * 1.0);
        int j = 1;
        for(int i = 1; i <= n; i++){
            sum[i] = 0;
            atag[i] = 0;
            scanf("%lld", &a[i]);
            sum[j] += a[i];
            if(i % SZ == 0) j++;
        }
        while(m--){
            char op[4];
            int x, y;
            LL c;
            scanf("%s %d %d", op, &x, &y);
            if(op[0] == 'Q'){
                printf("%lld
", query(x, y));
            }else{
                scanf("%lld", &c);
                update(x, y, c);
            }
        }
    }    
    return 0;
}

Paint The Wall

HDU - 4391

参考：https://www.cnblogs.com/swm8023/archive/2012/09/23/2698780.html

带懒标记的分块

// lazy 分块
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;

int c[maxn], ov[maxn];
map<int, int> mp[maxn];
int n, m, SZ;

void pushdown(int id){
    if(ov[id] != -1){
        for(int i = id * SZ; i < min((id + 1) * SZ, n); i++) c[i] = ov[id];
        mp[id].clear();
        mp[id][ov[id]] = min((id + 1) * SZ, n) - id * SZ;
        ov[id] = -1;
    }
}

void update(int l, int r, int z){
    int pl = l / SZ, pr = r / SZ;
    pushdown(pl);
    for(int i = l; i <= min(r, (pl + 1) * SZ - 1); i++){
        mp[pl][c[i]]--;
        c[i] = z;
        mp[pl][z]++;
    }
    if(pl != pr){
        pushdown(pr);
        for(int i = pr * SZ; i <= r; i++){
            mp[pr][c[i]]--;
            c[i] = z;
            mp[pr][z]++;
        } 
    }
    for(int i = pl + 1; i < pr; i++) ov[i] = z;
}
int query(int l, int r, int z){
    int res = 0;
    int pl = l / SZ, pr = r / SZ;
    pushdown(pl);
    for(int i = l; i <= min(r, (pl + 1) * SZ - 1); i++) if(c[i] == z) res++;
    if(pl != pr) {
        pushdown(pr);
        for(int i = pr * SZ; i <= r; i++) if(c[i] == z) res++;
    }
    for(int i = pl + 1; i < pr; i++){
        if(ov[i] == -1){
            if(mp[i].find(z) != mp[i].end()) res += mp[i][z];
        }else if(ov[i] == z) res += min((i + 1) * SZ, n) - i * SZ;
    } 
    return res;
}
int main(){
    while(scanf("%d %d", &n, &m) != EOF){
        SZ = sqrt(n * 1.0);
        memset(ov, -1, sizeof ov);
        for(int i = 0; i < n; i++) mp[i].clear();
        for(int i = 0; i < n; i++){
            scanf("%d", &c[i]);
            mp[i / SZ][c[i]]++; 
        }
        while(m--){
            int op, l, r, z;
            scanf("%d %d %d %d", &op, &l, &r, &z);
            if(op == 1){
                update(l, r, z);
            }else{
                printf("%d
", query(l, r, z));
            }
        }
    }
    return 0;
}