分块思想

CF348http://codeforces.com/contest/348/problem/C

题意：一个n个元素的序列，m个下标集合，q个操作，两种操作1：给你一个集合，求序列中下标在集合中的元素的sum；2：给你一个集合，把序列中下标在集合中的元素都加X;  1<= n,m,q <= 10^5;  并且题目中明确说明所有集合的元素个数小于10^5；

分析：http://codeforces.com/blog/entry/9031?locale=en

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
typedef long long LL;
const int N = 100000+10;
const int SZ = 315;
vector<int> g[N];
int id[N],id_sz;
int n,m,q;
LL a[N];
int ins[N][N/SZ+10];
bool inset[N/SZ+10][N];
LL sum[N/SZ+10],tar[N/SZ+10];
void solve(){
    LL an;
    while (q--) {
        char s[5];
        int v,x;
        scanf("%s",s);
        if (s[0] == '?') {
            scanf("%d",&v);
            v--;
            if (id[v] == -1) {
                an = 0;
                for (int i = 0; i < g[v].size(); i++) {
                    an += a[g[v][i]];
                }
                for (int i = 0; i < id_sz; i++) {
                    an += (LL)tar[i] * ins[v][i];
                }
                cout<<an<<endl;
            }else {
                an = sum[id[v]];
                for (int i = 0; i < id_sz; i++) {
                    an += (LL)tar[i] * ins[v][i];
                }
                cout<<an<<endl;
            }
        }else {
            scanf("%d%d",&v,&x);
            v--;
            if (id[v] == -1) {
                for (int i = 0; i < g[v].size(); i++) {
                    a[g[v][i]] += x;
                }
                for (int i = 0; i < id_sz; i++) {
                    sum[i] += ins[v][i] * x;
                }
            }else {
                tar[id[v]] += x;
            }
        }

    }
}
int main(){
    while (~scanf("%d%d%d",&n,&m,&q)) {
        for (int i = 0; i < n; i++) scanf("%lld",a+i);
        memset(id,-1,sizeof(id));
        id_sz = 0;
        for (int i = 0; i < m; i++) g[i].clear();
        for (int i = 0; i < m; i++) {
            int cnt; scanf("%d",&cnt);
            LL tmp = 0;
            while (cnt--) {
                int x; scanf("%d",&x); x--;
                g[i].push_back(x);
                tmp += a[x];
            }
            if (g[i].size() > SZ) {
                for (int j = 0; j < g[i].size(); j++)
                    inset[id_sz][g[i][j]] = 1;
                sum[id_sz] = tmp;
                id[i] = id_sz++;
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < g[i].size(); j++) {
                for (int k = 0; k < id_sz; k++) {
                    ins[i][k] += inset[k][g[i][j]];
                }
            }
        }
        solve();
    }
    return 0;
}
/*
5 3 5
5 -5 5 1 -4
2 1 2
4 2 1 4 5
2 2 5
? 2
+ 2 4
? 1
+ 2 1
? 1
*/