• codeforces 551E. GukiZ and GukiZiana 分块


    题目链接

    给出n个数, m个操作, 每个操作有两种, 将一段区间加上某个值, 或者询问一个k, a[i] = a[j] = k, 输出满足条件的最大的j-i, 如果没有输出-1。

    做法是将数组分块, 第一次做这种, 抄的codeforces上面的代码...

    #include<bits/stdc++.h>
    using namespace std;
    #define pb(x) push_back(x)
    #define ll long long
    #define mk(x, y) make_pair(x, y)
    #define lson l, m, rt<<1
    #define mem(a) memset(a, 0, sizeof(a))
    #define rson m+1, r, rt<<1|1
    #define mem1(a) memset(a, -1, sizeof(a))
    #define mem2(a) memset(a, 0x3f, sizeof(a))
    #define rep(i, a, n) for(int i = a; i<n; i++)
    #define ull unsigned long long
    typedef pair<int, int> pll;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int mod = 1e9+7;
    const int inf = 1061109567;
    const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
    multiset <int> s[1000];
    const int maxn = 5e5+5;
    int a[maxn], N, add[1005], n;
    void modify(int l, int r, int val) {
        for(int i = l; i<=r; i++) {
            if(i%N==0&&i+N-1<=r) {
                add[i/N] += val;
                if(add[i/N]>1e9)
                    add[i/N] = 1e9+1;
                i+=N-1;
            } else {
                s[i/N].erase(s[i/N].find(a[i]));
                a[i]+=val;
                if(a[i]>1e9)
                    a[i] = 1e9+1;
                s[i/N].insert(a[i]);
            }
        }
    }
    int query(int x) {
        int l = -1, r = -2;
        for(int i = 0; i<n; i++) {
            if(a[i] + add[i/N] == x) {
                l = i;
                break;
            }
            if(i%N==0&&i+N-1<n) {
                if(s[i/N].find(x-add[i/N])==s[i/N].end())
                    i+=N-1;
            }
        }
        for(int i = n-1; i>=0; i--) {
            if(a[i] + add[i/N] == x) {
                r = i;
                break;
            }
            if((1+i)%N == 0) {
                if(s[i/N].find(x-add[i/N]) == s[i/N].end())
                    i = i-N+1;
            }
        }
        return r-l;
    }
    int main()
    {
        int m, cnt = 0;
        cin>>n>>m;
        N = sqrt(n*1.0);
        for(int i = 0; i<n; i++) {
            scanf("%d", &a[i]);
            if(i&&i%N==0)
                cnt++;
            s[cnt].insert(a[i]);
        }
        while(m--) {
            int sign, l, r, val;
            scanf("%d", &sign);
            if(sign == 1) {
                scanf("%d%d%d", &l, &r, &val);
                l--, r--;
                modify(l, r, val);
            } else {
                scanf("%d", &val);
                cout<<query(val)<<endl;
            }
        }
        return 0;
    }

     3.10 update 重新写了一发。

    #include <iostream>
    #include <vector>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <map>
    #include <set>
    #include <string>
    #include <queue>
    #include <stack>
    #include <bitset>
    using namespace std;
    #define pb(x) push_back(x)
    #define ll long long
    #define mk(x, y) make_pair(x, y)
    #define lson l, m, rt<<1
    #define mem(a) memset(a, 0, sizeof(a))
    #define rson m+1, r, rt<<1|1
    #define mem1(a) memset(a, -1, sizeof(a))
    #define mem2(a) memset(a, 0x3f, sizeof(a))
    #define rep(i, n, a) for(int i = a; i<n; i++)
    #define fi first
    #define se second
    typedef pair<ll, int> pll;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int mod = 1e9+7;
    const int inf = 1e18;
    const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
    const int maxn = 5e5+5;
    vector<pll> b[800];
    ll a[maxn], c[maxn], l[800], r[800], add[800], belong[maxn];
    int n, cnt, block;
    int ansx, ansy;
    void rebuild(int x) {
        b[x].clear();
        for(int i = l[x]; i<=r[x]; i++) {
            b[x].pb(mk(a[i], i));
        }
        sort(b[x].begin(), b[x].end());
    }
    void update(int x, int y, int val) {
        if(belong[x] == belong[y]) {
            for(int i = x; i<=y; i++) {
                a[i]+=val;
            }
            rebuild(belong[x]);
            return ;
        }
        for(int i = x; i<=r[belong[x]]; i++) {
            a[i]+=val;
        }
        for(int i = l[belong[y]]; i<=y; i++) {
            a[i]+=val;
        }
        for(int i = belong[x]+1; i<belong[y]; i++) {
            add[i] += val;
        }
        rebuild(belong[x]);
        rebuild(belong[y]);
    }
    void bin(int x, ll val) {
        auto it = lower_bound(b[x].begin(), b[x].end(), mk(val, 0));
        if(it == b[x].end())
            return ;
        if(it->first == val) {
            ansx = min(ansx, it->second);
        }
        it = lower_bound(b[x].begin(), b[x].end(), mk(val+1, 0));
        if(it == b[x].begin())
            return ;
        it--;
        if(it -> first == val) {
            ansy = max(ansy, it->second);
        }
    }
    void query(ll val) {
        for(int i = 1; i<=cnt; i++) {
            ll tmp = val-add[i];
            if(tmp<0)
                continue;
            bin(i, tmp);
        }
    }
    int main()
    {
        int q, x, y, z, sign;
        cin>>n>>q;
        block = sqrt(n*1.0);
        for(int i = 1; i<=n; i++) {
            scanf("%I64d", &a[i]);
            belong[i] = (i-1)/block+1;
        }
        if(n%block)
            cnt = n/block+1;
        else
            cnt = n/block;
        for(int i = 1; i<=cnt; i++) {
            l[i] = (i-1)*block+1;
            r[i] = i*block;
        }
        r[cnt] = n;
        for(int i = 1; i<=cnt; i++)
            rebuild(i);
        while(q--) {
            scanf("%d", &sign);
            if(sign == 1) {
                scanf("%d%d%d", &x, &y, &z);
                update(x, y, z);
            } else {
                scanf("%d", &x);
                ansx = inf, ansy = -inf;
                query(x);
                if(ansx==inf) {
                    puts("-1");
                } else {
                    printf("%d
    ", ansy-ansx);
                }
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/yohaha/p/5071329.html
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