HDU 4391 Paint The Wall（分块的区间维护）

题意：给出几个操作，把l-r赋值为z，询问l-r有几个z，其中z < INT_MAX

思路：因为z很大，所以很难直接用线段树去维护。这里可以使用分块来解决。我们可以让每个块用map去储存map[i]的个数，用类似线段树的lazy标记来给整个块更新，当需要对块内某些数操作时再pushdown。

注意一下不要随意开辟map的空间，在计算区间的z的个数时应采用

if(b[i].num.find(z) != b[i].num.end()) ans += b[i].num[z];

减少空间开辟。

代码：

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 100000 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
struct Block{
    map<int, int> num;
    int lazy, L, R;
}b[1000];
int belong[maxn], a[maxn], block, sz;
int n, m;
void init(){
    block = sqrt(n * 1.0);
    for(int i = 0; i < n; i++)
        belong[i] = i / block + 1;
    sz = belong[n - 1];
    for(int i = 1; i <= sz; i++){
        b[i].lazy = -1;
        b[i].L = (i - 1) * block;
        b[i].R = min(b[i].L + block - 1, n - 1);
        b[i].num.clear();
        for(int j = b[i].L; j <= b[i].R; j++){
            b[i].num[a[j]]++;
        }
    }
}
void push_down(int x){
    b[x].num.clear();
    for(int i = b[x].L; i <= b[x].R; i++)
        a[i] = b[x].lazy;
    b[x].num[b[x].lazy] = b[x].R - b[x].L + 1;
    b[x].lazy = -1;
}
void update(int ll, int rr, int z){
    int l = belong[ll], r = belong[rr];
    int L, R;
    if(l == r){
        if(b[l].lazy != -1) push_down(l);
        for(int i = ll; i <= rr; i++){
            b[l].num[a[i]]--;
            b[l].num[z]++;
            a[i] = z;
        }
    }
    else{
        L = ll, R = b[l].R;
        if(b[l].lazy != -1) push_down(l);
        for(int i = L; i <= R; i++){
            b[l].num[a[i]]--;
            b[l].num[z]++;
            a[i] = z;
        }

        L = l + 1, R = r - 1;
        for(int i = L; i <= R; i++)
            b[i].lazy = z;

        L = b[r].L, R = rr;
        if(b[r].lazy != -1) push_down(r);
        for(int i = L; i <= R; i++){
            b[r].num[a[i]]--;
            b[r].num[z]++;
            a[i] = z;
        }
    }
}
int query(int ll, int rr, int z){
    int l = belong[ll], r = belong[rr];
    int L, R, ans = 0;

    if(l == r){
        if(b[l].lazy != -1) push_down(l);
        for(int i = ll; i <= rr; i++){
            if(a[i] == z) ans++;
        }
        return ans;
    }
    else{
        L = ll, R = b[l].R;
        if(b[l].lazy != -1) push_down(l);
        for(int i = L; i <= R; i++){
            if(a[i] == z) ans++;
        }

        L = l + 1, R = r - 1;
        for(int i = L; i <= R; i++){
            if(b[i].lazy != -1){
                if(b[i].lazy == z) ans += b[i].R - b[i].L + 1;
            }
            else{
                if(b[i].num.find(z) != b[i].num.end()) ans += b[i].num[z];  //不开辟新空间
            }
        }

        L = b[r].L, R = rr;
        if(b[r].lazy != -1) push_down(r);
        for(int i = L; i <= R; i++){
            if(a[i] == z) ans++;
        }
        return ans;
    }
}
int main(){
    while(~scanf("%d%d", &n, &m)){
        for(int i = 0; i < n; i++){
            scanf("%d", &a[i]);
        }
        init();
        while(m--){
            int o, l, r, z;
            scanf("%d%d%d%d", &o, &l, &r, &z);
            if(o == 1){
                update(l, r, z);
            }
            else{
                printf("%d\n", query(l, r, z));
            }
        }
    }
    return 0;
}