Codeforces 348C Subset Sums 分块思想

Subset Sums

把所有集合分成大于sqrt(n)和小于sqrt(n)的集合， 处理出每两个集合有交集是多大之后就可以在sqrt(n)的复杂内完成一次操作。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

const int B = 350;

int n, m, q;
LL a[N];
int cnt[307][N];
int cnt2[307][307];

LL sum[N], addval[N];
int to[N];
int fto[N];
vector<int> S[N];
vector<int> big, sml;
vector<int> vc[N];

int main() {
    scanf("%d%d%d", &n, &m, &q);
    for(int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    for(int i = 1; i <= m; i++) {
        int num; scanf("%d", &num);
        S[i].resize(num);
        for(int j = 0; j < num; j++) scanf("%d", &S[i][j]);
        if(num >= B) to[i]= SZ(big), fto[SZ(big)] = i, big.push_back(i);
        else to[i] = SZ(sml), fto[SZ(sml)] = i, sml.push_back(i);
    }
    for(int i = 0; i < SZ(big); i++) {
        for(auto& x : S[big[i]]) {
            vc[x].push_back(i);
            sum[fto[i]] += a[x];
        }
    }
    for(int i = 0; i < SZ(sml); i++) {
        for(auto& x : S[sml[i]]) {
            for(auto& y : vc[x]) {
                cnt[y][i]++;
            }
        }
    }
    for(int i = 0; i < SZ(big); i++) {
        for(auto& x : S[big[i]]) {
            for(auto& y : vc[x]) {
                if(i < y) cnt2[i][y]++, cnt2[y][i]++;
            }
        }
    }
    while(q--) {
        char op[3];
        scanf("%s", op);
        if(op[0] == '+') {
            int k, x; scanf("%d%d", &k, &x);
            if(SZ(S[k]) >= B) {
                sum[k] += 1LL * SZ(S[k]) * x;
                addval[k] += x;
                for(int i = 0; i < SZ(big); i++) {
                    if(to[k] == i) continue;
                    sum[fto[i]] += 1LL * x * cnt2[to[k]][i];
                }
            } else {
                for(auto& id : S[k]) a[id] += x;
                for(int i = 0; i < SZ(big); i++)
                    sum[fto[i]] += 1LL * cnt[i][to[k]] * x;
            }
        } else {
            int k; scanf("%d", &k);
            if(SZ(S[k]) >= B) {
                printf("%lld\n", sum[k]);
            } else {
                LL ans = 0;
                for(auto& x : S[k]) ans += a[x];
                for(int i = 0; i < SZ(big); i++)
                    ans += addval[fto[i]] * cnt[i][to[k]];
                printf("%lld\n", ans);
            }
        }
    }
    return 0;
}
/*
*/