題目連接：http://acm.hdu.edu.cn/showproblem.php?pid=3874

Necklace

Time Limit: 15000/5000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3929????Accepted Submission(s): 1296

Problem Description Mery has a beautiful necklace. The necklace is made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful value look the same, so if two or more balls have the same beautiful value, we just count it once. We define the beautiful value of some interval [x,y] as F(x,y). F(x,y) is calculated as the sum of the beautiful value from the xth ball to the yth ball and the same value is ONLY COUNTED ONCE. For example, if the necklace is 1 1 1 2 3 1, we have F(1,3)=1, F(2,4)=3, F(2,6)=6.

Now Mery thinks the necklace is too long. She plans to take some continuous part of the necklace to build a new one. She wants to know each of the beautiful value of M continuous parts of the necklace. She will give you M intervals [L,R] (1<=L<=R<=N) and you must tell her F(L,R) of them.

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Input The first line is T(T<=10), representing the number of test cases.
??For each case, the first line is a number N,1 <=N <=50000, indicating the number of the magic balls. The second line contains N non-negative integer numbers not greater 1000000, representing the beautiful value of the N balls. The third line has a number M, 1 <=M <=200000, meaning the nunber of the queries. Each of the next M lines contains L and R, the query.

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Output For each query, output a line contains an integer number, representing the result of the query.

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Sample Input 2 6 1 2 3 4 3 5 3 1 2 3 5 2 6 6 1 1 1 2 3 5 3 1 1 2 4 3 5

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Sample Output 3 7 14 1 3 6 題意：　求區(qū)間不重復(fù)元素和 題解：　詢問(wèn)是５個(gè)０　所以肯定要找一種可以掃描一次得出結(jié)果不會(huì)重復(fù)計(jì)算的算法，這里介紹一種巧妙的方法：　離線操作，所謂離線就是將所有的詢問(wèn)都讀如后，根據(jù)需要進(jìn)行排序，這里一定要記錄一個(gè)id來(lái)保存輸入的順序，然后一個(gè)ans數(shù)組按照輸入順序儲(chǔ)存答案，然后從左到右的掃描一邊將所有的以當(dāng)前掃描的點(diǎn)為終止點(diǎn)的答案都保存起來(lái)，最后按順序輸出ans 數(shù)組即可 對(duì)于這個(gè)題：一般按照右端點(diǎn)排序，因?yàn)橛叶它c(diǎn)代表這一個(gè)查詢的結(jié)束，所以按右端點(diǎn)排序有特殊的意義，因?yàn)槊看我コ貜?fù)的數(shù)字，可以考慮設(shè)一個(gè)last數(shù)組，標(biāo)記其在之前是否出現(xiàn)過(guò)，如果出現(xiàn)過(guò)的化，將之前出現(xiàn)的地方的這個(gè)值更改成０　然后last數(shù)組的當(dāng)前值更新成當(dāng)前的位置，這樣樹(shù)狀數(shù)組中存放的值就肯定沒(méi)有重復(fù)元素了。 這種離線的思想很重要，一定要利用好將所有的數(shù)據(jù)排序后每次可以將相同的結(jié)尾的值一起算出來(lái)的性質(zhì)，指針只用掃描一邊，每向后掃描后都將以這個(gè)值結(jié)尾的所有結(jié)果都處理出來(lái)。離線最大的好處是在依次向后更新last數(shù)組的時(shí)候前面的與其相關(guān)的詢問(wèn)已經(jīng)處理過(guò)了，所以去除前面的值，保留后面的不會(huì)影響前面的結(jié)果，也不會(huì)影響后面的計(jì)算。如果不離線，last 數(shù)組如果這樣更新的話，會(huì)影響到再次涉及前面的區(qū)間的和 代碼： 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 #define N 1000005 6 #define M 50005 7 #define numq 200005 8 #define ll long long 9 int Last[N]; 10 ll shusz[M]; 11 int gaga[M]; 12 ll ans[numq]; 13 struct Q { 14 int l ; 15 int r; 16 int id; 17 bool operator < (const Q a) const 18 { 19 return r<a.r; 20 } 21 }qq[numq]; 22 int lb(int i) 23 { 24 return i&(-i); 25 } 26 void add(int j , int t) 27 { 28 for(int i =j ;i < M ;i+=lb(i)) 29 { 30 shusz[i]+=t; 31 } 32 } 33 ll sum (int x) 34 { 35 ll ans = 0 ; 36 for(int i = x ; i > 0 ; i-=lb(i)) 37 { 38 ans+=shusz[i]; 39 } 40 return ans; 41 } 42 ll sum(int x , int y) 43 { 44 ll ans = sum(y)-sum(x-1);//注意是x-1 45 return ans; 46 } 47 int main() 48 { 49 int T ; 50 scanf("%d",&T); 51 for(int i =0 ;i < T ; i++) 52 { 53 int n; 54 scanf("%d",&n); 55 int tm; 56 for(int j = 1 ; j <= n ; j++) 57 { 58 scanf("%d",&tm); 59 gaga[j] = tm; 60 } 61 int m ; 62 scanf("%d",&m); 63 for(int j = 1 ;j <= m ;j++) 64 { 65 int l , r ; 66 scanf("%d%d",&l,&r); 67 qq[j].l = l ; 68 qq[j].r = r; 69 qq[j].id = j; 70 } 71 memset(Last,-1,sizeof(Last)); 72 memset(ans,0,sizeof(ans)); 73 memset(shusz,0,sizeof(shusz)); 74 sort(qq+1,qq+m+1); 75 int cur = 1;//記錄掃描到第幾個(gè)詢問(wèn) 76 for(int j = 1 ; j <= n ; j++)//掃描n個(gè)點(diǎn) 77 { 78 if(Last[gaga[j]] != -1) 79 add(Last[gaga[j]], -gaga[j]); 80 Last[gaga[j]] = j; 81 add(j, gaga[j]); 82 while(j == qq[cur].r) 83 { 84 ans[qq[cur].id] = sum(qq[cur].l, qq[cur].r); 85 cur++; 86 } 87 } 88 for(int j = 1; j <= m; j++) 89 printf("%lld\n", ans[j]); 90 } 91 return 0 ; 92 }

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轉(zhuǎn)載于:https://www.cnblogs.com/shanyr/p/4725462.html

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