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
1 #include<cstdio> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cmath> 5 #include<iostream> 6 #include<algorithm> 7 #include<vector> 8 #include<queue> 9 using namespace std; 10 typedef long long LL; 11 const int N = 100000+10; 12 const int SZ = 315; 13 vector<int> g[N]; 14 int id[N],id_sz; 15 int n,m,q; 16 LL a[N]; 17 int ins[N][N/SZ+10]; 18 bool inset[N/SZ+10][N]; 19 LL sum[N/SZ+10],tar[N/SZ+10]; 20 void solve(){ 21 LL an; 22 while (q--) { 23 char s[5]; 24 int v,x; 25 scanf("%s",s); 26 if (s[0] == '?') { 27 scanf("%d",&v); 28 v--; 29 if (id[v] == -1) { 30 an = 0; 31 for (int i = 0; i < g[v].size(); i++) { 32 an += a[g[v][i]]; 33 } 34 for (int i = 0; i < id_sz; i++) { 35 an += (LL)tar[i] * ins[v][i]; 36 } 37 cout<<an<<endl; 38 }else { 39 an = sum[id[v]]; 40 for (int i = 0; i < id_sz; i++) { 41 an += (LL)tar[i] * ins[v][i]; 42 } 43 cout<<an<<endl; 44 } 45 }else { 46 scanf("%d%d",&v,&x); 47 v--; 48 if (id[v] == -1) { 49 for (int i = 0; i < g[v].size(); i++) { 50 a[g[v][i]] += x; 51 } 52 for (int i = 0; i < id_sz; i++) { 53 sum[i] += ins[v][i] * x; 54 } 55 }else { 56 tar[id[v]] += x; 57 } 58 } 59 60 } 61 } 62 int main(){ 63 while (~scanf("%d%d%d",&n,&m,&q)) { 64 for (int i = 0; i < n; i++) scanf("%lld",a+i); 65 memset(id,-1,sizeof(id)); 66 id_sz = 0; 67 for (int i = 0; i < m; i++) g[i].clear(); 68 for (int i = 0; i < m; i++) { 69 int cnt; scanf("%d",&cnt); 70 LL tmp = 0; 71 while (cnt--) { 72 int x; scanf("%d",&x); x--; 73 g[i].push_back(x); 74 tmp += a[x]; 75 } 76 if (g[i].size() > SZ) { 77 for (int j = 0; j < g[i].size(); j++) 78 inset[id_sz][g[i][j]] = 1; 79 sum[id_sz] = tmp; 80 id[i] = id_sz++; 81 } 82 } 83 for (int i = 0; i < n; i++) { 84 for (int j = 0; j < g[i].size(); j++) { 85 for (int k = 0; k < id_sz; k++) { 86 ins[i][k] += inset[k][g[i][j]]; 87 } 88 } 89 } 90 solve(); 91 } 92 return 0; 93 } 94 /* 95 5 3 5 96 5 -5 5 1 -4 97 2 1 2 98 4 2 1 4 5 99 2 2 5 100 ? 2 101 + 2 4 102 ? 1 103 + 2 1 104 ? 1 105 */
