[hdu5213]容斥原理+莫队算法

浏览数：17 / 时间：2015年06月08日

题意：给一个序列a，以及K，有Q个询问，每个询问四个数，L,R,U,V, 求L<=i<=R,U<=j<=V，a[i]+a[j]=K的(i, j)对数（题目保证了L <= R < U <= V）。

思路：首先用容斥原理把询问变为i,j同区间，记f(A,B)为答案，‘+‘为区间的并，A=[L,R],B=[U,V],C=[u+1,v-1]，则f(A,B) = f(A+B+C,A+B+C)+f(C,C)-f(A+C,A+C)-f(B+C,B+C)。令g(L,R) = f([L,R],[L,R]) = f(A,A)。对于g函数区间变化为1时，可以做到o(1)的维护。用莫队算法的知识把询问排个序，按顺序维护当前区间的答案，同时更新最后答案。

总结：莫队算法并不是某一具体问题的解法，而是一种通用算法，对于任意无修改的区间统计问题，都可以用莫队算法试试。对于所有询问(L,R)，复杂度等于sigma(|Li - Li-1| + |Ri - Ri-1|)*o(x)，o(x)是区间变化为1时的维护代价，而前面的sigma用莫队算法的知识可以降低到o(m*sqrt(n))。

  1 #pragma comment(linker, "/STACK:10240000,10240000")
  2 
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <cstring>
  8 #include <map>
  9 #include <queue>
 10 #include <deque>
 11 #include <cmath>
 12 #include <vector>
 13 #include <ctime>
 14 #include <cctype>
 15 #include <set>
 16 #include <bitset>
 17 #include <functional>
 18 #include <numeric>
 19 #include <stdexcept>
 20 #include <utility>
 21 
 22 using namespace std;
 23 
 24 #define mem0(a) memset(a, 0, sizeof(a))
 25 #define mem_1(a) memset(a, -1, sizeof(a))
 26 #define lson l, m, rt << 1
 27 #define rson m + 1, r, rt << 1 | 1
 28 #define define_m int m = (l + r) >> 1
 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 32 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 33 #define all(a) (a).begin(), (a).end()
 34 #define lowbit(x) ((x) & (-(x)))
 35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 38 #define pchr(a) putchar(a)
 39 #define pstr(a) printf("%s", a)
 40 #define sstr(a) scanf("%s", a)
 41 #define sint(a) scanf("%d", &a)
 42 #define sint2(a, b) scanf("%d%d", &a, &b)
 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 44 #define pint(a) printf("%d\n", a)
 45 #define test_print1(a) cout << "var1 = " << a << endl
 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 48 #define mp(a, b) make_pair(a, b)
 49 #define pb(a) push_back(a)
 50 
 51 typedef long long LL;
 52 typedef pair<int, int> pii;
 53 typedef vector<int> vi;
 54 
 55 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
 56 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
 57 const int maxn = 3e4 + 7;
 58 const int md = 10007;
 59 const int inf = 1e9 + 7;
 60 const LL inf_L = 1e18 + 7;
 61 const double pi = acos(-1.0);
 62 const double eps = 1e-6;
 63 
 64 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
 65 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 66 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 67 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 68 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 69 int make_id(int x, int y, int n) { return x * n + y; }
 70 
 71 struct Node {
 72     int L, R, id, kth;
 73     constructInt4(Node, L, R, id, kth);
 74 };
 75 Node node[maxn * 4];
 76 
 77 int block;
 78 int a[maxn], ans[maxn], cnt[maxn];
 79 
 80 bool cmp(const Node a, const Node b) {
 81     int lb = a.L / block, rb = b.L / block;
 82     return lb < rb || lb == rb && a.R < b.R;
 83 }
 84 
 85 int main() {
 86     //freopen("in.txt", "r", stdin);
 87     int n, m, k;
 88     while (cin >> n) {
 89         cin >> k;
 90         rep_up0(i, n) {
 91             sint(a[i]);
 92         }
 93         cin >> m;
 94         rep_up0(i, m) {
 95             int L, R, U, V;
 96             scanf("%d %d %d %d", &L, &R, &U, &V);
 97             L --; R --; U --; V --;
 98             node[i * 4] = Node(L, V, i, 0);
 99             node[i * 4 + 1] = Node(R + 1, U - 1, i, 1);
100             node[i * 4 + 2] = Node(L, U - 1, i, 2);
101             node[i * 4 + 3] = Node(R + 1, V, i, 3);
102         }
103 
104         block = (int)sqrt(n + 0.5);
105         sort(node, node + 4 * m, cmp);
106 
107         int cur_ans = 0, L = 0, R = -1;
108         mem0(cnt);
109         mem0(ans);
110 
111         rep_up0(i, 4 * m) {
112             Node query = node[i];
113             while (L < query.L) {
114                 cnt[a[L]] --;
115                 if (k > a[L] && k - a[L] <= n) cur_ans -= cnt[k - a[L]];
116                 L ++;
117             }
118             while (R > query.R) {
119                 cnt[a[R]] --;
120                 if (k > a[R] && k - a[R] <= n) cur_ans -= cnt[k - a[R]];
121                 R --;
122             }
123             while (L > query.L) {
124                 L --;
125                 if (k > a[L] && k - a[L] <= n) cur_ans += cnt[k - a[L]];
126                 cnt[a[L]] ++;
127             }
128             while (R < query.R) {
129                 R ++;
130                 if (k > a[R] && k - a[R] <= n) cur_ans += cnt[k - a[R]];
131                 cnt[a[R]] ++;
132             }
133             if (query.kth < 2) ans[query.id] += cur_ans;
134             else ans[query.id] -= cur_ans;
135         }
136 
137         rep_up0(i, m) {
138             printf("%d\n", ans[i]);
139         }
140     }
141     return 0;
142 }