hdu 3627 Giant For

Giant For

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2404    Accepted Submission(s): 536

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

5add 48 1add 25 69add 88 52remove 25 69add 23 8910add 47 23find 66 83find 27 73add 84 97find 10 58remove 47 23add 41 89remove 41 89find 65 68add 25 410

 

Sample Output

Case 1:Case 2:-1-184 9784 97
把矩阵中的点，离散化成一维的点，然后用线段树乱搞。
#include <stdio.h>#include <string.h>#include <iostream>#include <string>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#include <stdlib.h>#include <algorithm>using namespace std;typedef long long ll;#define rep(i,s,t) for(int i=s;i<t;i++)#define red(i,s,t) for(int i=s-1;i>=t;i--)#define ree(i,now) for(int i=head[now];i!=-1;i=edge[i].next)#define clr(a,v) memset(a,v,sizeof a)#define L t<<1#define R t<<1|1#define MID int mid=(l+r)>>1//#define max(a,b) (a<b?b:a)//#define min(a,b) (a<b?a:b)#define lowbit(x) (x&(-x))#define SQR(a) ((a)*(a))inline int input(){    int ret=0;bool isN=0;char c=getchar();    while(c<'0' || c>'9'){        if(c=='-') isN=1;        c=getchar();    }    while(c>='0' && c<='9'){        ret=ret*10+c-'0';        c=getchar();    }    return isN?-ret:ret;}inline void output(int x){    if(x<0){        putchar('-');x=-x;    }    int len=0,data[11];    while(x){        data[len++]=x%10;x/=10;    }    if(!len)    data[len++]=0;    while(len--)        putchar(data[len]+48);    putchar('\n');}const int MAXN=200005;struct op{    char op[10];    int a,b;}q[MAXN];int n,to[MAXN],k;map< pair<int,int>,int>mp;map< pair<int,int>,int>::iterator it;int val[MAXN<<2];inline void modefiy(int t,int l,int r,int x,int v){    if(l==r) val[t]=v;    else{        MID;        if(x<=mid) modefiy(L,l,mid,x,v);        else modefiy(R,mid+1,r,x,v);        val[t]=max(val[L],val[R]);    }}inline pair<int,int> query(int t,int l,int r,int x,int y){    if(val[t]<=y) return make_pair(-1,-1);    if(to[r]<=x) return  make_pair(-1,-1);    if(l==r) return make_pair(to[l],val[t]);    MID;    pair<int,int> p=query(L,l,mid,x,y);    if(p.second==-1) return query(R,mid+1,r,x,y);    return p;}int main(){    int ca=1;    while(n=input(),n){        if(ca>1) puts("");        printf("Case %d:\n",ca++);        k=0,mp.clear();        rep(i,0,n){            scanf("%s%d%d",q[i].op,&q[i].a,&q[i].b);            mp[make_pair(q[i].a,q[i].b)]=1;        }        for(it=mp.begin();it!=mp.end();it++){            (*it).second=(++k);            to[k]=(*it).first.first;        }        clr(val,-1);        rep(i,0,n){            if(q[i].op[0]=='a'){                modefiy(1,1,k,mp[make_pair(q[i].a,q[i].b)],q[i].b);            }            else if(q[i].op[0]=='r'){                modefiy(1,1,k,mp[make_pair(q[i].a,q[i].b)],-1);            }            else{                pair<int,int> ans=query(1,1,k,q[i].a,q[i].b);                if(ans.second==-1) puts("-1");                else printf("%d %d\n",ans.first,ans.second);            }        }    }    return 0;}