Fzu 2059 MM (and check set + sort insert)

problem 2059 MMaccept:109 submit:484
Time limit:1000 mSec Memory limit:32768 KB problem Description

There is a array contain N (1<n<=100000) numbers. Now give you M (1<m<10000) query.

Every query would be:

1 X:ask longest substring which every number no less than X

2 y X:change the a[y] to X. There is at the very change times.

For each ask can you tell me the length of longest substring.

Input

There is multiple tests.

Each test first line contain the integer numbers n m,second line contain n integer numbers.

Next M lines Each line would be:

1 X:ask longest substring which every number no less than X

2 y X:change the a[y] to X. There is at the very change times.

0 < N <= 100000, 0 < M <= 10000, -1000000000 <= a[i] <= 1000000000

Outputeach ask output the length of longest substring. Sample INPUT5 2 3 2 3 3 sample Output3110 test instructions: Give a sequence, two operations, the first operation is to enter an X, find a substring satisfies each element's value is not less than X and the maximum length, output this maximum value, the first Two operations are input x and Y, modify a[y]=x, the second operation up to only 10 times thought: O (nlogn) preprocessing, O (logn) query. Because the modification operation only 10 times, so each modification is directly violent pre-treatment, and then consider the situation without modifying the operation. Preprocessing methods: Sorting the sequence values, and then from large to small insert, with and check the set to maintain the maximum length of the inserted point can be connected, in this process to maintain a max, the maximum value is the answer.

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespaceStd;typedefLong LongLl;typedef pair<int,int>PII;Const intINF =1e9;Const DoubleEPS = 1e-6;Const intN =100010;intCAS =1;struct_node{intPos,val,ans; FriendBOOL operator< (Const_node &a,Const_node &b) {returnA.val <B.val; }};intn,m;intFa[n],sum[n],b[n],mxval;_node A[n];intFindintx) {    returnfa[x]==x?x:fa[x]=find (Fa[x]);}voidUnintUintv) {u=find (u), v=Find (v); Fa[u]=v; SUM[V]+=sum[u];}voidPre () {memset (sum,0,sizeof(sum)); Mxval=-inf-Ten;  for(intI=0; i<n;i++) fa[i]=i;  for(intI=0; i<n;i++) {A[i].val=b[i],a[i].pos=i; if(A[i].val > Mxval) mxval =A[i].val; } sort (A,a+N); intMX =0;  for(inti=n-1; i>=0; i--) {Sum[a[i].pos]=1; if(sum[a[i].pos-1]) un (a[i].pos-1, A[i].pos); if(sum[a[i].pos+1]) un (a[i].pos+1, A[i].pos); if(MX < Sum[find (a[i].pos)]) MX =Sum[fa[a[i].pos]]; A[i].ans=MX; }}intSolveintx)    {_node T; T.val=x; intid = lower_bound (a,a+n,t)-A; returnA[id].ans;}voidrun () { for(intI=0; i<n;i++) scanf ("%d", B +i);    Pre (); intX,y,op;  while(m--) {scanf ("%d",&op); if(op==1) {scanf ("%d",&x); if(X>mxval) puts ("0"); Elseprintf"%d\n", Solve (x)); }        Else{scanf ("%d%d",&y,&x); B[y-1]=x;        Pre (); }    }}intMain () {#ifdef LOCAL freopen ("Case.txt","R", stdin); #endif     while(SCANF ("%d%d", &n,&m)! =EOF) run (); return 0;}

Fzu 2059 MM (and check set + sort insert)