HDU-3874 Necklace 线段树+离线
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 }
