Problem 2136 Fetch Candy---fuoj (line tree + maintenance)

http://acm.fzu.edu.cn/problem.php?pid=2136

The main idea: to give you n a bag with candy in each bag, and then you can take a continuous bag of one interval at a time, then carry the bag with the maximum number of candies in it.

Find the interval length from 1 to n the number of candies you can take away the worst case scenario is the minimum value of the maximum value for all intervals.

Analysis: If you look up from a small start, he will be the maximum of the interval whenever he finds the interval he can cover, so we can query this point every time.

#include <stdio.h>#include<string.h>#include<math.h>#include<algorithm>#include<ctype.h>#include<iostream>#include<map>using namespacestd;#defineINF 1e9+7#defineLson r<<1|1#defineRson r<<1#defineN 101000intAns[n];structnode{intL,r,la,lb,max;///LA is the number of consecutive left of this interval, LB is to the right and Max is the largest continuous sequence of intervals.    intmid () {return(L+R)/2; }    intLen () {return(r-l+1); }}a[n*Ten];structnode{intx, y;} P[n];intCMP (Node c,node d) {returnc.x<d.x;}voidBuildtree (intRintLintR) {A[R].L=L; A[R].R=R; A[r].la= a[r].lb = A[r]. Max =0; if(L = =R)return;    Buildtree (Lson, L, A[r].mid ()); Buildtree (Rson, A[r].mid ()+1, R);}voidQurry (intRintXinty) {    if(A[R].L = =A[R].R) {a[r].la= a[r].lb = A[r]. Max =1; return; }    if(Y >A[r].mid ())    Qurry (Rson, x, y); ElseQurry (Lson, x, y); A[R]. Max= Max (A[lson]. Max, Max (A[rson). Max, a[lson].lb+a[rson].la)); A[r].la=a[lson].la; a[r].lb=a[rson].lb; if(a[lson].la = =A[lson].len ()) a[r].la+=a[rson].la; if(a[rson].lb = =A[rson].len ()) a[r].lb+=a[lson].lb;}intMain () {intT; scanf ("%d", &T);  while(T--)    {        intN; scanf ("%d", &N); memset (P,0,sizeof(P));  for(intI=0; i<n; i++) {scanf ("%d", &p[i].x); P[i].y=i; } buildtree (1,0, N-1); Sort (p, p+N, CMP); intm=1; memset (ans,0,sizeof(ans));  for(intI=0; i<n; i++) {Qurry (1, p[i].x, P[I].Y);///Query the maximum sequence of y points            inttmp=a[1].            Max;  while(M <=tmp) Ans[m++]=p[i].x; }         for(intI=1; i<=n; i++) printf ("%d\n", Ans[i]); }    return 0;}

Problem 2136 Fetch Candy---fuoj (line tree + maintenance)