HDU5249-KPI（线段树+离散化）

Posted 2020-09-23 cxchanpin

好久没写线段树都不知道怎么写了。。。

很easy的线段树二分问题

#include<cstdio>
#include<set>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
#define lson (pos<<1)
#define rson (pos<<1|1)
const int maxn = 10005;
int n,Case = 1;
char op[10];
int  Hash[maxn];
int  cnt;
struct In{
    int op,v;
}in[maxn];
int HASH(int x){
    return lower_bound(Hash,Hash + cnt,x) - Hash + 1;
}
int sum[maxn << 2];
void pushup(int pos){
    sum[pos] = sum[lson] + sum[rson];
    return;
}
void update(int L,int R,int m,int pos,int value){
    if(L == R){
        sum[pos] = value;
        return;
    }
    int mid = (L + R) >> 1;
    if(m <= mid)
        update(L,mid,m,lson,value);
    else
        update(mid + 1,R,m,rson,value);
    pushup(pos);
    return;
}
int query(int L,int R,int value,int pos){
    if(L == R)
        return L;
    int mid = (L + R) >> 1;
    if(sum[lson] >= value)
        return query(L,mid,value,lson);
    else
        return query(mid + 1,R,value - sum[lson],rson);
}
int main(){
    while(scanf("%d",&n) != EOF){
        printf("Case #%d:\n",Case++);
        cnt = 0;
        queue<int>q;
        memset(sum,0,sizeof(sum));
        for(int i = 0; i < n; i++){
            scanf("%s",op);
            if(op[0] == ‘i‘){
                in[i].op = 1;
                scanf("%d",&in[i].v);
                Hash[cnt++] = in[i].v;
            }
            else if(op[0] == ‘o‘)
                in[i].op = 2;
            else if(op[0] == ‘q‘)
                in[i].op = 3;
        }
        sort(Hash,Hash + cnt);
        cnt = unique(Hash,Hash + cnt) - Hash;
        for(int i = 0; i < n; i++){
            if(in[i].op == 1){
                int e = HASH(in[i].v);
                q.push(e);
                update(1,cnt,e,1,1);
            }
            else if(in[i].op == 2){
                int e = q.front(); q.pop();
                update(1,cnt,e,1,0);
            }
            else if(in[i].op == 3){
                int f  = sum[1];
                int fx = f / 2 + 1;
                int pos = query(1,cnt,fx,1);
                printf("%d\n",Hash[pos - 1]);
            }
        }
    }
    return 0;
}
/*
10
in 1
in 2
in 3
in 4
in 5
o
q
o
q
q
*/