标签：output more which input name use oid iostream after
个人心得：线段树的延迟标志确实是减少了很多时间，思想比较简单，但是实现得时候和建立延迟的时候比较麻烦。
按照我的一些理解，就是更新时找到完全覆盖的区间时，更新延迟标志，不再往下更新，但此时父节点啥的都会更新，但是
递归思想到了这里还是会回去，所以在程序末尾进行往上更新就好了，同时，在查询的时候有延迟标志时要下放，
但是注意此时不会影响父节点的值，因为在更新延迟标志的时候就已经回溯把父节点更新了。
题目：
InputThe input contains servel test cases. The first line is the case number. In each test case:
The first line contains just one number k( 1 ≤ k ≤ 1000 ) and Q( 1 ≤ Q ≤ 100000 )
The following lines, each line contains two integers a and b, ( 1 ≤ a < b ≤ 1000000 ), indicate a query.
Huge Input, scanf recommanded.OutputFor each test case, output three lines:
Output the case number in the first line.
If the ith query can be satisfied, output i. i starting from 1. output an blank-space after each number.
Output a blank line after each test case.Sample Input
1 3 6 1 6 1 6 3 4 1 5 1 2 2 4
Sample Output
Case 1: 1 2 3 5
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<iomanip> 6 #include<algorithm> 7 using namespace std; 8 #define inf 1<<29 9 #define nu 4000005 10 #define maxnum 1000000 11 int n,k; 12 int maxelem; 13 int book[100005]; 14 int flag=0; 15 typedef struct 16 { 17 int left,right; 18 int sum; 19 int jf; 20 int mid(){ 21 return (left+right)/2; 22 } 23 24 }Tree; 25 Tree tree[nu]; 26 void buildtree(int root,int l,int r){ 27 tree[root].left=l,tree[root].right=r; 28 tree[root].sum=tree[root].jf=0; 29 if(l!=r){ 30 buildtree(root*2+1,l,(l+r)/2); 31 buildtree(root*2+2,(l+r)/2+1,r); 32 } 33 } 34 void updown(int root) 35 { 36 37 tree[root*2+1].jf+=tree[root].jf; 38 tree[root*2+2].jf+=tree[root].jf; 39 tree[root*2+1].sum+=tree[root].jf; 40 tree[root*2+2].sum+=tree[root].jf; 41 tree[root].jf=0; 42 } 43 void upset(int root) 44 { 45 tree[root].sum=max(tree[root*2+1].sum,tree[root*2+2].sum); 46 } 47 int checktree(int root,int l,int r) 48 { 49 if(tree[root].left==l&&tree[root].right==r) 50 { 51 return tree[root].sum; 52 } 53 int mid=tree[root].mid(); 54 if(tree[root].jf) updown(root); 55 if(r<=mid) 56 return checktree(root*2+1,l,r); 57 else if(l>mid) 58 return checktree(root*2+2,l,r); 59 else 60 { 61 return max(checktree(root*2+1,l,mid),checktree(root*2+2,mid+1,r)); 62 } 63 } 64 void inserttree(int root,int l,int r){ 65 if(tree[root].left==l&&tree[root].right==r) 66 { 67 tree[root].jf+=1; 68 tree[root].sum+=1; 69 return ; 70 } 71 int mid=tree[root].mid(); 72 if(tree[root].jf) updown(root); 73 if(r<=mid) 74 inserttree(root*2+1,l,r); 75 else if(l>mid) 76 inserttree(root*2+2,l,r); 77 else 78 { 79 inserttree(root*2+1,l,mid); 80 inserttree(root*2+2,mid+1,r); 81 } 82 upset(root); 83 } 84 int main() 85 { 86 int t,j; 87 scanf("%d",&t); 88 for(j=1;j<=t;j++){ 89 scanf("%d%d",&k,&n); 90 buildtree(0,1,maxnum); 91 flag=0; 92 memset(book,0,sizeof(book)); 93 for(int i=1;i<=n;i++){ 94 int a,b; 95 scanf("%d%d",&a,&b); 96 b--; 97 if(checktree(0,a,b)<k){ 98 book[flag++]=i; 99 inserttree(0,a,b); 100 } 101 } 102 printf("Case %d:\n",j); 103 for(int p=0;p<flag;p++) 104 { 105 printf("%d ",book[p]); 106 } 107 printf("\n\n"); 108 } 109 return 0; 110 }
标签：output more which input name use oid iostream after
原文地址：http://www.cnblogs.com/blvt/p/7895308.html
