Algorithm training operation Lattice

Source: Internet
Author: User

Algorithm training operation lattice time limit: 1.0s memory limit: 256.0MBProblem description

There are n squares, left-to-right in a row, numbered 1-n.

There are 3 types of operation for m operations:

1. Modify the weights of a lattice,

2. To seek a continuous lattice of weights and

3. Ask for the maximum value of a continuous lattice.

For each of the 2, 3 operation output you have to find the results.

Input format

The first row of 2 integers n,m.

The next line of n integers represents the initial weights of the n lattices.

Next m line, 3 integers per line p,x,y,p represents the type of operation, p=1 when the weight of the modified lattice X is y,p=2 when the weighted value of the interval [x, y] is calculated, and the p=3 indicates the maximum weight of the interval [x, y] lattice.

Output format

There are several lines, the number of rows equals the total number of p=2 or 3 operations.

1 integers per line, corresponding to the result of each p=2 or 3 operation.

Sample Input 4 3
1 2 3 4
2 1 3
1 4 3
3 1 4 Sample output 6
3 Data size and conventions

For 20% of data n <= 100,m <= 200.

For 50% of data n <= 5000,m <= 5000.

For 100% data 1 <= n <= 100000,m <= 100000,0 <= lattice weights <= 10000.

#include <iostream>#include<cstdio>#defineMAXN 500004structnode{intLeft,right; intnum;} tree[3*MAXN];intmaxv[2*MAXN];int  out=0;intMaxintXinty) {    if(x>y)returnx; Else       returny;}voidBuildintLeftintRightinti) {Tree[i].left=Left ; Tree[i].right=Right ; Tree[i].num=0; if(tree[i].left==tree[i].right)return ; intMid= (left+right)/2; Build (Left,mid,2*i); Build (Mid+1, right,2*i+1);}voidInsertintIdintIintj) {      if(tree[id].left==i&&tree[id].right==i) {tree[id].num=J; Maxv[id]=J; }      if(tree[id].left==tree[id].right)return ; if(i>tree[id].right)return; if(i<tree[id].left)return; intMid= (tree[id].left+tree[id].right)/2; if(i<=mid) Insert (ID*2, i,j); ElseInsert (ID*2+1, i,j); Tree[id].num=tree[2*id].num+tree[2*id+1].num; Maxv[id]=max (maxv[id*2],maxv[id*2+1]);}voidSumintIdintIintj) {    intMid= (tree[id].left+tree[id].right)/2; if(tree[id].left==i&&tree[id].right==j) { out+=Tree[id].num; return ; }     if(j<=mid) sum (ID*2, i,j); Else if(i>mid) sum (ID*2+1, i,j); Else{sum (id*2, I,mid); SUM (ID*2+1, mid+1, J); }}intQueryintIdintLintR) {      intMid= (tree[id].left+tree[id].right)/2; intans=0; if(l==tree[id].left&&r==tree[id].right) {                                  returnMaxv[id]; }      if(r<=mid) ans=max (Ans,query (id*2, L,r)); Else if(l>mid) ans=max (Ans,query (id*2+1, L,r)); Else{ans=max (Ans,query (id*2, L,mid)); Ans=max (Ans,query (id*2+1, mid+1, R)); }     returnans;}intMain () {intI,p,x,y; intN,m,u;  while(~ scanf ("%d%d",&n,&m) {Build (1N1);  for(i=1; i<=n;i++) {scanf ("%d",&u); Insert (1, I,u); }     for(i=1; i<=m;i++) {scanf ("%d%d%d",&p,&x,&y); if(p==1) Insert (1, x, y); if(p==2) {sum (1, x, y); printf ("%d\n", out);  out=0; }         if(p==3) {printf ("%d\n", Query (1, x, y)); }     }   }     return 0;}

Algorithm training operation Lattice

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