(线段树 + 离散化 ) hdu 3627 Giant For

Giant For

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1397    Accepted Submission(s): 261

Problem Description It is known to us all that YY and LMY are mathematics lovers. They like to find and solve interesting mathematic problems together. Now LMY designs a game for matrix. There is a large matrix whose rows and columns are both not more than 1000000000. And there are three operations for the matrix:
1) add: Mark an element in the matrix. It is guaranteed that the element has not been marked before.
2) remove: Delete an element’s mark. It is guaranteed that the element has been marked before.
3) find: For a given element’s row and column, return a marked element’s row and column, where the marked element’s row and column are larger than the given element’s row and column respectively. If there are multiple solutions, return the element whose row is the smallest; and if there are still multiple solutions, return the element whose column is the smallest. If there is no solution, return -1.
LMY lets YY develop a program to solve the problem. Could you also develop a program to solve the problem?
 

 

Input The input consists of multiple test cases. For each test case, the first line contains only one integer n. n ≤ 200000. Each of the next n lines describes an operation. There is a blank line between two consecutive test cases.
End of input is indicated by a line containing a zero.
 

 

Output Start each test case with "Case #:" on a single line, where # is the case number starting from 1. For each “find” operation, output the result. The format is showed as sample output. There is a blank line between two consecutive test cases.
 

 

Sample Input

5
add 48 1
add 25 69
add 88 52
remove 25 69
add 23 89

10
add 47 23
find 66 83
find 27 73
add 84 97
find 10 58
remove 47 23
add 41 89
remove 41 89
find 65 68
add 25 41
0

Sample Output

Case 1:

Case 2:
-1
-1
84 97
84 97

Source The 35th ACM/ICPC Asia Regional Tianjin Site —— Online Contest 2010天津网络预选赛1007题目 ps：当时没学线段树，一看数据这么大就蒙了，现在学了线段树就做了起来。效率还不错1548ms AC 线段树的资料：请参考http://blog.csdn.net/Hashmat/archive/2010/09/28/5911369.aspx /*题目大意：不超过*1000000000的矩阵，三个操作。add 插入(xi,yi)；remove 移除 (xi,yi)；find 查找(>xi,>yi)的最小值题目解答：n=1000000000 太大超过 int 的范围：op_num(操作数) = 200000 也就是最多有op_num对数值 所以离散成 hash[op_num]然后再用线段树统计Problem : 3627 ( Giant For ) Judge Status : AcceptedRunId : 3007530 Language : C++ Author : wawadimuCode Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta*/#include<iostream>#include<algorithm>using namespace std;#define N 200010//操作数组char op[N][7];//原数组，排序数组，离散后数组struct node{int row,col;}res[N],tmp[N],hash[N];// 线段树struct tr{int l,r,max_col;}tree[3*N];//按行升序排列int cmp( const node &a , const node &b){if( a.row != b.row ) return a.row < b.row;else return a.col < b.col ;}//在区间[L,R]内寻找p的次序int query(node p , int L , int R);//建立线段树void built(int l, int r , int num);//更新区间[cnt,cnt]void insert(node p, int cnt , int num);//移除区间[cnt,cnt];void remove(node p, int cnt ,int num);//查找区间[L,R]内比L大的节点int find(node p , int L ,int R , int num);int main(){freopen("3627.txt","r",stdin);int n;int i,j,k;int c=1;while(scanf("%d",&n)!=EOF && n!=0){if( c != 1) printf("/n");printf("Case %d:/n",c++);for(i=1; i<=n; i++){scanf("%s%d%d",op[i], &res[i].row, &res[i].col);tmp[i].row=res[i].row; tmp[i].col=res[i].col;}sort(tmp+1,tmp+n+1,cmp);//排序k=0;tmp[0].col=-N; tmp[0].row=-N;for(i=1; i<=n; i++)//离散{if( tmp[i].row == tmp[i-1].row && tmp[i].col == tmp[i-1].col) continue;else{ //cout<<tmp[i].row<<"->"<<tmp[i].col<<" ";hash[++k].col=tmp[i].col;hash[k].row=tmp[i].row;}}//cout<<k<<endl;built(1 , k , 1 );for(i=1; i<=n; i++){int cnt=query( res[i] , 1 , k); //查找res[i]在离散后的数组的位置hash[cnt]//cout<<cnt<<" ";if( strcmp( op[i] ,"add") == 0) insert( res[i] , cnt ,1 );else if(strcmp( op[i] , "remove") == 0) remove( res[i] , cnt ,1 );else { int ans=find( res[i] , cnt , k , 1 );//找到返回区间值[ans,ans]，否则返回-1if( ans ==-1 ) printf("-1/n");else printf("%d %d/n",hash[ans].row,hash[ans].col);}}}return 0;}void built( int l , int r , int num)//建立单点的区间{tree[num].l=l; tree[num].r= r; tree[num].max_col=0;//cout<<num<<" :"<<l<<"->"<<r<<" ";if(l<r){int m=( l + r ) >> 1;built( l , m , 2*num );built( m+1 , r, 2*num+1 );//m+1}}int query(node p , int L , int R){int l=L , r=R ;int t = p.row;////由于可以出现行重叠的情况，所以查找比他row大的数字，然后反推比较确定值的位置//while( l <= r){int mid=( l + r ) >> 1;if( t >= hash[mid].row )l= mid + 1;else r= mid -1;}//反向比较，确定位置while(--l){if(hash[l] .col == p.col )break;}return l;}void insert( node p, int cnt , int num){if( tree[num].l == tree[num].r )//叶子节点，插入{//cout<<tree[num].l<<"->"<<p.col<<" ";tree[num].max_col = p.col;return ;}//cout<<num<<" ";int mid=( tree[num].l + tree[num].r ) >> 1;if( cnt <= mid ) insert( p , cnt , 2*num );else insert( p , cnt , 2*num + 1);//父亲节点保存子树y的最大值，有利于降低查找的效率tree[num].max_col = tree[2*num].max_col > tree[2*num+1].max_col ? tree[2*num].max_col : tree[2*num+1].max_col;}void remove( node p, int cnt , int num){if( tree[num].l == tree[num].r ){//cout<<tree[num].l<<"->"<<p.col<<" ";tree[num].max_col = 0;return ;}//cout<<cnt<<num<<" ";int mid=( tree[num].l + tree[num].r ) >> 1;if( cnt <= mid ) remove( p , cnt , 2*num );else remove( p , cnt , 2*num + 1);}int find( node p , int L , int R ,int num){int tmp;if( tree[num].max_col < p.col ) return -1;if( tree[num].l == tree[num].r ){//tree[num].max_col > p.col && tree[num].l > L cout<<tree[num].l<<">"<<L<<" ";if( hash[tree[num].l].row > p.row && hash[tree[num].r].col > p.col )//这里比较实际值的大小不是离散值大小，因为这里WA了.{return tree[num].l;}else return -1;}int mid = ( tree[num].l + tree[num].r ) >> 1;////如果覆盖两个子区域，先找左区间是否有符合的值，然后查找右区间的情况//if( L < mid ) {//cout<<2*num<<" "; tmp = find( p, L , R, 2*num ); if( tmp != -1 ) return tmp;}if( R >= mid ){//cout<<2*num+1<<" ";tmp=find( p, L , R, 2*num + 1 );if( tmp != -1 ) return tmp;}return -1;}