HDU 3874 Necklace 区间查询的离线操作

题目: http://acm.hdu.edu.cn/showproblem.php?pid=3874

对需要查询的区间按右端点排序,然后从左到右依次加入序列中的元素,同时更新,更新的方法是,把上一次出现a[i]值的点变为0,这一次a[i]值的点(即 i)变为a[i],这样保证了前i个元素中只存在一个等于a[i]值得元素,那为什么这样不会影响后面的查询呢?

因为是处理到i点,则把右边界等于a[i]的查询处理掉,剩下的待查询的区间右边界在i点之后,如果左边界在i之前,那么也会包含i点,也就包含了i点的值,由于i以前没有等于a[i]的点,所以只包含了一个这样的值,如果左边界在i之后,前面的操作对它就没影响了。也可以这样理解,当前处理到i点,如果后面的待查询区间的左边界要包含上一个值为a[i]的点,那么它必须也包含了i点,所以i之前等于a[i]的点完全可以舍弃-----------------原来先排序的处理方式还有个专业名字叫=======离线操作

话不多说,看代码:

  1 /**********************************************
  2 ***    Problem:
  3 ***    Author:        JKL
  4 ***    University:    CSUST
  5 ***    Team:          __Dream
  6 ***    Email:          1451108308@QQ.COM
  7 ***    My Blog:        http://www.cnblogs.com/jklongint/
  8 ***********************************************/
  9 //===================================================
 10 #include <iostream>
 11 #include <fstream>
 12 #include <sstream>
 13 #include <iomanip>
 14 #include <cstdio>
 15 #include <cstdlib>
 16 #include <cmath>
 17 #include <cassert>
 18 #include <numeric>
 19 #include <ctime>
 20 #include <algorithm>
 21 #include <cstring>
 22 #include <string>
 23 #include <vector>
 24 #include <queue>
 25 #include <map>
 26 #include <stack>
 27 #include <list>
 28 #include <set>
 29 #include <bitset>
 30 #include <deque>
 31 using namespace std;
 32 //---------------------------------------------------
 33 #define mem(a,b) memset(a,b,sizeof(a))
 34 #define GO cout<<"HelloWorld!"<<endl
 35 #define Case(x) cout<<"Case "<<x<<":"
 36 #define foru(i,n) for(int i=1; i <= n; i++)
 37 #define ford(i,n) for(int i = n; i >= 1; i--)
 38  #define fin freopen("input.txt","r",stdin);
 39  #define fout freopen("output.txt","w",stdout)
 40 #define lson  l, m, rt << 1
 41 #define rson  m + 1, r, rt << 1 | 1
 42 
 43 #define sqr(a)  ((a)*(a))
 44 #define abs(a) ((a>0)?(a):-(a))
 45 #define pii pair<int,int>
 46 
 47 #define fmax(a,b) max(a,b)
 48 #define fmin(a,b) min(a,b)
 49 #define fmax3(a,b,c)  (fmax(a,fmax(a,b)))
 50 #define fmin3(a,b,c)  (fmin(a,fmin(a,b)))
 51 
 52 #define sfi(x) scanf("%d",&x)
 53 #define sfL(x) scanf("%I64d",&x)
 54 #define sfc(x) scanf("%c",&x)
 55 #define sfd(x) scanf("%lf",&x)
 56 #define sfs(x) scanf("%s",x)
 57 #define sfii(a,b) scanf("%d%d",&a,&b)
 58 #define sfLL(a,b) scanf("%I64d%I64d",&a,&b)
 59 #define sfcc(a,b) scanf("%c%c",&a,&b)
 60 #define sfdd(a,b) scanf("%lf%lf",&a,&b)
 61 #define sfss(a,b) scanf("%s%s",a,b)
 62 
 63 #define pfi(x) printf("%d",x)
 64 #define pfL(x) printf("%I64d",x)
 65 #define pfs(x) printf("%s",x)
 66 #define pfd(x) printf("%lf",x)
 67 #define pfc(x) print("%c",x)
 68 #define newLine pfs("
")
 69 #define space pfs(" ")
 70 
 71 //--------------------------------------------------------
 72 typedef __int64 LL;
 73 typedef unsigned long long ULL;
 74 //typedef __int64 __LL;
 75 typedef unsigned __int64 __ULL;
 76 
 77 typedef vector<int> vi;
 78 typedef vector<LL> vL;
 79 typedef vector<string> vs;
 80 typedef set<int> si;
 81 typedef map<int,int> mii;
 82 typedef map<LL,LL> mLL;
 83 typedef map<string,int> msi;
 84 typedef map<char,int> mci;
 85 //--------------------------------------------------------
 86 const int dx[4]={1,-1,0,0};
 87 const int dy[4]={0,0,1,-1};
 88  const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 89  const int N6=1000006;
 90  const int N5=100006;
 91  const int N4=10006;
 92  const int N3=1006;
 93  const int N2=106;
 94  const int N=210009;
 95  const int MOD=1000000007;
 96  const LL LMAX=0x7fffffffffffffff;
 97  const LL IMAX=0x3fffffff;
 98  const double PI=3.14159265359;
 99 //--------------------------------------------------------
100 template< class T > T gcd(T a, T b) { return (b != 0 ? gcd<T>(b, a%b) : a); }
101 template< class T > T lcm(T a, T b) { return (a / gcd<T>(a, b) * b); }
102 
103 //------------------------------------------------------------
104 struct TreeNode{
105     LL  sum;
106 };
107 struct Node{
108     int l, r, id;
109 };
110 //=================================================================
111 TreeNode tree[N << 2];
112 Node node[N];
113 int  a[N], last[1000009];
114 LL ans[N];
115 int cmp(Node i, Node j)
116 {
117     return i.r < j.r ;
118 }
119 void PushUP(int rt)
120 {
121     tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
122 }
123 void update(int p, int x, int l , int r, int rt)
124 {
125     if(l == r){
126         tree[rt].sum += x;
127         return;
128     }
129     int m = (l + r) >> 1;
130     if(p <= m)update(p, x, lson);
131     else update(p, x, rson);
132     PushUP(rt);
133 }
134 LL query(int L, int R, int l, int r, int rt)
135 {
136     if(L <= l && R >= r){
137         return tree[rt].sum;
138     }
139     int m = (l + r) >> 1;
140     LL res = 0;
141     if(L <= m) res += query(L, R, lson);
142     if(R > m) res += query(L, R, rson);
143     return res;
144 }
145 void build(int l, int r, int rt)
146 {
147     if(l == r){
148         tree[rt].sum = 0;
149         return;
150     }
151     int m = (l + r) >> 1;
152     build(lson);
153     build(rson);
154     PushUP(rt);
155 }
156 int main()
157 {
158     //fin;//fout;//freopen("input.txt","r",stdin);
159     int n, m, T;
160     cin >> T;
161     while(T--){
162         cin >> n ;
163         foru(i, n)sfi(a[i]);
164         cin >> m;
165         foru(i, m)sfii(node[i].l, node[i].r),node[i].id = i;
166         sort(node + 1, node + 1 + m, cmp);
167         build(1, n, 1);
168         int np = 1;
169         mem(last, 0);
170         foru(i, n){
171             if(last[a[i]])update(last[a[i]], -a[i], 1, n, 1);
172             update(i, a[i], 1, n, 1);
173             while(node[np].r == i && np <= m){
174                 ans[node[np].id] = query(node[np].l, node[np].r, 1, n, 1);
175                 np++;
176             }
177             last[a[i]] = i;
178         }
179         foru(i, m)pfL(ans[i]),newLine;
180     }
181     return 0;
182 }
View Code
原文地址:https://www.cnblogs.com/jklongint/p/3891397.html