Bestcoder round #65 && hdu 5592 ZYB ' s premutation segment tree

Time limit:2000/1000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 175 Accepted Submission (s): 74

Problem DescriptionZYB Has a premutationP, but he is remeber the reverse log of each prefix of the premutation,now he ask you
Restore the premutation.

Pair(i,J)(i<j) is considered as a reverse log if ai>aJ is matched.

Inputin The first line there is the number of testcases T.

For each teatcase:

In the first line there is one numberN.

In the next line there isNNumbersAi , describe the number of the reverse logs of each prefix,

The input is correct.

1≤T≤5,1≤N≤50000

Outputfor each testcase,print the ans.

Sample Input130 1 2

Sample OUTPUT3 1 2

Sourcebestcoder Round #65

Idea: It is easy to think k = (A[i]-a[i-1] + 1) is the number of the original sequence I the left of a larger number, can be calculated from right to left

Using the line segment Tree implementation: Find the sequence of K positive number corresponding to the position of the subscript, did not find a, is equivalent to the position of the numbers into negative O (log)

#include <cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<cstdlib>#include<queue>#include<vector>#include<map>#include<Set>#include<iostream>#defineLson L, M, RT << 1#defineRson m + 1, R, RT << 1|1using namespacestd;Const intINF =0x3f3f3f3f; typedefLong Longll;Const intN =50005;intNum[n <<2], A[n], ans[n];voidUpintRT) {Num[rt] = Num[rt <<1] + num[rt <<1|1]; }voidBuildintLintRintRT) {    if(L = =R) {Num[rt]=1; return ; }    intm = (L + r) >>1;    Build (Lson);    Build (Rson); Up (RT);}voidUpdateintLintRintRtintp) {    if(L = = P && r = =p) {Num[rt]--; return ; }    intm = (L + r) >>1; if(P <=m) Update (Lson, p); ElseUpdate (Rson, p); Up (RT);}intPos;void Get(intLintRintRtintx) {    if(L = =r) {POS=l; return; }    intm = (L + r) >>1; if(Num[rt <<1] < X)Get(Rson, X-num[rt <<1]); Else Get(Lson, x);}intMain () {int_; scanf"%d", &_);  while(_ --)    {        intN scanf"%d", &N);  for(inti =1; I <= N; ++i) scanf ("%d", &A[i]); Build (1N1); intc =N;  for(inti = n; I >=2; --i) {intD = C-(A[i]-a[i-1]); Get(1N1, D); Update (1N1, POS); //cout << pos << Endl;Ans[i] =POS; C--; }        Get(1N1,1); ans[1] =POS;  for(inti =1; I < n; ++i) printf ("%d", Ans[i]); printf ("%d\n", Ans[n]); }    return 0;}

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Bestcoder round #65 && hdu 5592 ZYB ' s premutation segment tree