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HDU-3874 Necklace 线段树+离线

news 来源：原创 2024/4/25 6:30:37

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

　　比较简单的题，题意也好懂。

　　先O(n)求每个数左边第一次出现的与他相同的数的位置l[i]。对询问按照y从小大排序，然后按照从左到右的顺序来跟新点，当前点为i，那么删掉l[i]，加入点i，然后遇到询问求和。

  1 //STATUS:C++_AC_2593MS_10024KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef __int64 LL;
 34 typedef unsigned __int64 ULL;
 35 //const
 36 const int N=50010,M=200010;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=95041567,STA=8000010;
 39 const LL LNF=1LL<<60;
 40 const double EPS=1e-8;
 41 const double OO=1e15;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 struct Node{
 59     int a,b,id;
 60     bool operator < (const Node& a)const {
 61         return b<a.b;
 62     }
 63 }q[M];
 64 LL sum[N<<2],ans[M];
 65 int num[N],l[N],la[1000010];
 66 int T,n,m;
 67 
 68 void update(int l,int r,int rt,int w,int val)
 69 {
 70     if(l==r){
 71         sum[rt]=val;
 72         return;
 73     }
 74     int mid=(l+r)>>1;
 75     if(w<=mid)update(lson,w,val);
 76     else update(rson,w,val);
 77     sum[rt]=sum[rt<<1]+sum[rt<<1|1];
 78 }
 79 
 80 LL query(int l,int r,int rt,int L,int R)
 81 {
 82     if(L<=l && r<=R){
 83         return sum[rt];
 84     }
 85     int mid=(l+r)>>1;
 86     LL ret=0;
 87     if(L<=mid)ret+=query(lson,L,R);
 88     if(R>mid)ret+=query(rson,L,R);
 89     return ret;
 90 }
 91 
 92 int main()
 93 {
 94   //  freopen("in.txt","r",stdin);
 95     int i,j,k;
 96     scanf("%d",&T);
 97     while(T--)
 98     {
 99         scanf("%d",&n);
100         mem(la,0);
101         for(i=1;i<=n;i++){
102             scanf("%d",&num[i]);
103             l[i]=la[num[i]];
104             la[num[i]]=i;
105         }
106         scanf("%d",&m);
107         for(i=0;i<m;i++){
108             scanf("%d%d",&q[i].a,&q[i].b);
109             q[i].id=i;
110         }
111         sort(q,q+m);
112         mem(sum,0);k=0;
113         for(i=1;i<=n;i++){
114             if(l[i]){
115                 update(1,n,1,l[i],0);
116             }
117             update(1,n,1,i,num[i]);
118             for(;q[k].b==i && k<m;k++){
119                 ans[q[k].id]=query(1,n,1,q[k].a,q[k].b);
120             }
121         }
122         for(i=0;i<m;i++){
123             printf("%I64d\n",ans[i]);
124         }
125     }
126     return 0;
127 }